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Max-Albert Knus
Quadratic Forms,
Cli�ord Algebras and Spinors
IMECC-UNICAMP
1988
Max-Albert Knus
ETHMathematikR�amistr. 101CH{8092 Z�urichSwitzerland
AMS Subject Classi�cation (1985 Revision): 11E04, 11E16, 11E20, 11E39,11E57, 11E81, 11E88, 15A63, 15A66, 15A90, 20G15, 20G45
Foreword
These notes grew out of a course given at UNICAMP for two months between August and October1987. One aim of the course was to present the basic theory of quadratic forms and Cli�ord alge-bras over a �eld in a characteristic free way. In even dimension we restrict attention to nonsingularforms. In odd dimension nonsingularity implies that the characteristic must be di�erent from 2.An elegant way to avoid this restriction is to work with 1
2{regular forms. Such a form is the or-
thogonal sum of a 1{dimensional form and of a nonsingular form of even dimension. The 1
2{regular
forms behave almost as nicely as nonsingular forms and they are also de�ned in characteristic 2.A typical example in dimension 3 is given by the elements of reduced trace zero in a quaternionalgebra. The notion of 1
2{regularity is due to Kneser (see his G�ottingen lecture notes mentioned
in the bibliography at the end of these notes).Another purpose of the lectures was to give a detailled study of forms of low dimension, low
meaning � 6. The Cli�ord algebra gives much information for these forms. In particular theCli�ord algebra can be used to describe quite explicitly the spin group, the Lie algebra of theorthogonal group and the group of special similitudes of a form. In dimensions 4, 5, and 6 we haveapplied methods developed in joint papers with Parimala and Sridharan. These methods also givea di�erent approach to results of Dieudonn�e (Acta Math. 87, 1952). Moreover they do not dependon the characteristic of the �eld. The main tool is a reduced pfaÆan, introduced independentlyby Fr�ohlich, Jacobson and Tamagawa. Its role is very similar to the role of the reduced norm of aquaternion algebra in the study of forms of dimension 3 and 4. A general theory of the pfaÆan andapplications to quadratic forms of rank 6 over commutative rings can be found in the papers withParimala and Sridharan mentioned above. This theory of forms of rank 6 is related to the theoryof central simple algebras of dimension 16. In particular it yields new proofs of some theorems ofAlbert.
These notes have two appendices. In the �rst one we summarize, in a chart, all the provenresults on the structure of Cli�ord algebras in low dimensions. The second gives some concreteapplications. We introduce spinors, following Chevalley, and compute explicitly the spinor repre-sentations of four Cli�ord algebras occuring in physics: the Pauli algebra, the Minkowski algebra,the Majorana algebra and the Dirac algebra.
Even if this course deals with forms over �elds, it was sometimes convenient (and even nec-essary) to consider forms over commutative rings. Moreover readers, familiar with the techniquesof commutative algebra, will observe that the argument, used many times, to prove a result bypassing to some �eld extension, comes from descent theory.
We make the following conventions. The ground �eld is denoted by K. Vector spaces arealways �nite dimensional. Algebras are associative with 1 (if not explicitly mentioned) and K isidenti�ed with K � 1. Unadorned tensor products are taken over K. The group of units of analgebra is denoted by A�. Maps are written on the left, like functions. We use a unique numberingin each chapter for Examples, Lemmas, Propositions, Theorems and Corollaries. The bibliographyonly contains books and papers mentioned in the notes. We did not try systematically to givecredits for the results or the proofs used.
iv Foreword
I am very grateful to A. Paques, UNICAMP, who made my visit to Campinas possible. Heattended my course with patience, read the notes very carefully and made many suggestions (andcorrections). I also would like to thank P. H�ansli, who helped me a lot in correcting the �nalversion, R. Parimala and R. Sridharan, who read a preliminary version and Elda Mortari for herexcellent script with LATEX. Of course I am still responsible for any mistakes remaining. My visitto UNICAMP was �nanced by grants from FINEP and FAPESP. I thank these institutions andUNICAMP for their support.
Contents
1 Quadratic Forms 1
2 Central Simple Algebras 22
3 Involutions on Central Simple Algebras 35
4 The Cli�ord Algebra 56
5 Invariants of Quadratic Forms 67
6 Special Orthogonal Groups and Spin Groups 79
7 Quadratic Forms of Dimension 2 89
8 Quadratic Forms of Dimension 3 95
9 Quadratic Forms of Dimension 4 101
10 The PfaÆan 117
11 Quadratic Forms of Dimension 6 127
12 Quadratic Forms of Dimension 5 152
A A Chart of Results 162
B Spinors 166
v
Contents
Foreword
Chapter 1. Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2. Central Simple Algebras . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 3. Involutions on Central Simple Algebras . . . . . . . . 26
Chapter 4. The Cli�ord Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Chapter 5. Invariants of Quadratic Forms . . . . . . . . . . . . . . . . . 48
Chapter 6. Special Orthogonal Groups and Spin Groups . . . 57
Chapter 7. Quadratic Forms of Dimension 2 . . . . . . . . . . . . . . . 65
Chapter 8. Quadratic Forms of Dimension 3 . . . . . . . . . . . . . . . 69
Chapter 9. Quadratic Forms of Dimension 4 . . . . . . . . . . . . . . . 73
Chapter 10. The PfaÆan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Chapter 11. Quadratic Forms of Dimension 6 . . . . . . . . . . . . . . . 92
Chapter 12. Quadratic Forms of Dimension 5 . . . . . . . . . . . . . . . 110
Appendix A. A Chart of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Appendix B. Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Chapter 1
Quadratic Forms
This chapter gives basic facts about quadratic forms which hold over �elds in-
dependently of the characteristic. The proof of Theorem 10 (Witt cancellation) fol-
lows Wagner (see also the notes of Micali{Revoy). Useful references are the books of
Bourbaki, Chevalley, Lam, Scharlau, Baeza, Demazure{Gabriel and Micali{Revoy,
as well as the paper of Bass on Cli�ord algebras.
Let K be a �eld. A quadratic form is a pair (V; q), where V is a �nite dimen-
sional vector space over K and q is a map q : V ! K such that
1) q(�x) = �2q(x); � 2 K; x 2 V ;2) the map bq : V � V ! K de�ned by bq(x; y) = q(x + y) � q(x) � q(y) is
bilinear.
We call bq the polar of q.
Let b : V � V ! K be a symmetric bilinear form on V . The map
h : V ! V � = HomK(V;K); h(x) = b(x; �); x 2 V;
is called the adjoint of b. If we identify V with V �� through the map x 7! x��; x��(f) =
f(x); f 2 V �, we have h�(x) = b(�; x); where h� : V �� ! V � is the transpose of h,
i.e. h�(x��)(y) = x��(h(y)) = h(y)(x). Hence h� = h; since b is symmetric. We say
that b is nonsingular and that the pair (V; b) is a symmetric bilinear space if h is an
isomorphism. A quadratic form q : V ! K is called nonsingular and the pair (V; q)
is called a quadratic space if the polar bq is nonsingular. We denote its adjoint by
hq. If the characteristic of K is not equal to 2, the polar bq determines q, since
q(x) = 12bq(x; x); x 2 V;
2 1. Quadratic Forms
and the theory of quadratic spaces is identical with the theory of symmetric bilinear
spaces. The two theories are quite di�erent in characteristic 2. In these notes we
shall restrict to quadratic spaces.
A morphism of quadratic forms
' : (V; q)! (V 0; q0)
is a K{linear map ' : V ! V 0 such that q0('(x)) = q(x) for all x 2 V . Similarly a
morphism of bilinear forms
' : (V; b)! (V 0; b0)
is a K{linear map ' : V ! V 0 such that b0('(x); '(y)) = b(x; y) for all x; y 2 V .
A morphism ' : (V; b) ! (V 0; b0) (resp. (V; q) ! (V 0; q0)) must be injective if b
(resp. q) is nonsingular. Morphisms, which are isomorphisms, are called isometries.
The group of isometries of (V; b) (resp. (V; q)) is denoted by O(V; b) (resp. O(V; q))
and is called the orthogonal group of (V; b) (resp. (V; q)). An isometry of quadratic
forms induces an isometry of the associated polar forms but not conversely (if char
K = 2), see Example 2.
Let q : V ! K be a quadratic form and K � L a �eld extension. We can
extend the form q to a form qL : L V ! L by putting
qL(Xi
�i vi) =Xi
�2i q(vi) +Xi<j
�i�jbq(vi; vj):
The polar of qLis 1 bq and its adjoint is 1 hq : L V ! L V � = (L V )�. It
is usual to denote qLby 1 q and (LV; 1 q) by L (V; q). Obviously L (V; q)
is nonsingular if, and only if, (V; q) is nonsingular.
Let fe1; : : : eng be a basis of V and let
ai = q(ei); aij = bq(ei; ej); 1 � i; j � n:
The form q is determined by the ai and aij, since
q(x) =Xi
aix2i +
Xi<j
aijxixj for x =Xi
xiei:
1. Quadratic Forms 3
In matrix notation, we have q(x) = �t��, �t denoting the transpose of �, where
� =
1CCA and � =
0BBBB@a1 a12 � � � a1n0 a2 � � � a2n... 0
. . ....
0 � � � 0 an
1CCCCA :
Conversely, for any (n � n)-matrix �, we can de�ne a quadratic form q on V by
putting q(x) = �t�� for x =Pi xiei and � = (x1; : : : xn)
t.The matrix � is not
uniquely determined by q. We have
�t�� = �t�0�; 8� 2 Kn , � = �0 + ; alternating:
We recall that an alternating matrix is a matrix of the form
=
0BBBB@0 c12 � � � c1n
�c12 0 � � � c2n...
.... . .
...�c1n �c2n � � � 0
1CCCCAi.e. is antisymmetric and has zeroes on the diagonal. A matrix is alternating if
and only if = Æ� Æt for some matrix Æ. Thus, a basis fe1; : : : eng of V being �xed,
q is determined by the class [�] of � modulo alternating matrices and each such class
corresponds to a quadratic form. Moreover in each class [�] there is a triangular
matrix. We say that � and �0 are equivalent if [�] = [�0] i.e. if �� �0 is alternating.
If � is as above, the matrix � = (bq(ei; ej)) of the polar is
� =
0BBBB@2a1 a12 � � � a1na12 2a2 � � � a2n...
.... . .
...a1n a2n � � � 2an
1CCCCA = � + �t
and bq(x; y) = �t�� with � = (x1; : : : ; xn)t; � = (y1; : : : ; yn)
t if x =Pxiei and
y =Pyiei. Obviously � depends only on the class [�] of �. Let fe�1; : : : ; e�ng be the
dual basis of V �, i.e. e�i (ej) = Æij. We get
hq(ej) =Xi
aije�i
so that (V; q) is nonsingular if and only if det� 6= 0. We call det� the discriminant of
q with respect to the basis fe1; : : : ; eng and denote it by d(e1; : : : ; en). Let fe01; : : : ; e0ngbe another basis of V and let
ej =Xi
uije0i:
4 1. Quadratic Forms
Let � be the matrix (uij) and let � 0 = (bq(e0i; e
0j)). We have
� = �t� 0�:
Let K�2 be the set of squares in K�. The class of d(e1; : : : ; en) modulo K�2 does not
depend on the choice of the basis fe1; : : : ; eng. We denote this class by disc(q) and
call it the discriminant of q. Similarly, if ' : (V; q)! (V 0; q0) is an isometry and
'(ej) =Xi
uije0i
for bases fe1; : : : ; eng of V , resp. fe01; : : : ; e0ng of V 0, we have � = �t� 0�, putting � =
(bq(ei; ej)); � 0 = (bq0(e0i; e
0j)) and � = (uij). The condition � = �t� 0� is necessary
and suÆcient for the existence of an isometry of the bilinear spaces (V; bq); (V0; b0q)
but is not suÆcient for an isometry of the quadratic spaces (V; q) and (V 0; q0). If y =
'(x); x 2 V , let y = Pi yie
0i; x =
Pi xiei; � = (y1; : : : ; yn)
t and � = (x1; : : : ; xn)t,
so that � = ��. Assume that q(x) = �t�� and q0(y) = �t�0� for matrices �, resp. �0.
We have
�t�� = �t�0� = �t�t�0�� for all � 2 Kn:
Therefore � 2 GLn(K) de�nes an isometry if and only if [�] = [�t�0�]. It follows
that isometry classes of quadratic spaces are in 1� 1{correspondence with equiva-
lence classes [�] of matrices � 2Mn(K) such that � + �t is nonsingular.
Remark 1. The interpretation of a quadratic space as the class [�] of a matrix �
modulo alternating matrices was already applied in Klingenberg{Witt. This inter-
pretation has led to important generalizations of the notion of a quadratic space (see
the papers or books by Bak, Bass, Tits, Wall and Scharlau). We shall try to give an
idea of these generalizations by some examples. Let R be a ring with an involution
r 7! r, i.e. an automorphism of the additive group of R such that 1 = 1; rs = s r
and r = r for all r; s 2 R. Further let " = �1 and let � be an additive subgroup of
R such that
(1) fr � "r; r 2 Rg � � � fr 2 R j r = �"rg,(2) for all r 2 R; r�r � �.
1. Quadratic Forms 5
Further, let
�"n = f� 2Mn(R) j � =
0BBBB@�1 a12 � � � a1n
�"a12 �2. . .
�"a1n �n
1CCCCA ; aij 2 R; �i 2 �g:
For any matrix � = (aij), we put �� = (aij)
t. It follows from the properties (1) and
(2) of � that
2 �"n ) �� � 2 �"n for all � 2Mn(R):
We de�ne now an equivalence relation � � �0 on Mn(R) by
� � �0 , �� �0 2 �"n
and denote the equivalence class by [�]. We say that [�] is an (";�){quadratic
form on Rn and that [�] is nonsingular if � = � + "�� is invertible (observe that �
depends only on the class [�] of �). An isometry [�]! [�0] is by de�nition a matrix
� 2 GLn(R) such that [�] = [���0�]. We remark that an (";�){quadratic form on
Rn is not, as in the classical theory, a quadratic map Rn ! R. But we can associate
to [�] a map
q : Rn ! R=�; q(�) = class of ���� mod �:
We now consider di�erent special cases of (";�){quadratic forms. Let �min =
fr � " r; r 2 Rg and �max = fr 2 R j r = �" rg.1) R a �eld with trivial involution, " = 1 and � = �min: quadratic forms,
� = �max: symmetric bilinear forms.
2) R a �eld with trivial involution, " = �1 and � = �min: alternating forms,
� = �max: antisymmetric bilinear forms.
3) R a skew �eld with a nontrivial involution, " = 1 � = �min : even hermitian
forms, � = �max: hermitian forms.
More exotic cases are possible. We give two examples (without claiming that
the corresponding theories have applications!)
4) R a �eld of characteristic 2 with trivial involution, " = 1 (!) and � = (R2;+),
the additive group of squares in R.
5) R a commutative ring with trivial involution, " = �1 and � an ideal of R
6 1. Quadratic Forms
such that 2R � �.
We now give an important example of a quadratic space:
Example 2. Let V be a 2{dimensional vector space over K with a basis fe; fg. We
de�ne a quadratic form on V by
q(xe+ yf) = xy or q(�) = �t
0 10 0
!� for � =
xy
!:
The polar bq has matrix (0110), hence (V; q) is nonsingular. We observe that (V; q) '
(V; �q) for any � 2 K�, since 1 00 �
! 0 10 0
! 1 00 �
!=
0 �0 0
!:
The quadratic space (V; q) is called the hyperbolic plane and is denoted by H(K).
We have
O(V; bq) =
8<: x yu v
!2 GL2(K) j
x yu v
!t 0 11 0
! x yu v
!=
0 11 0
!9=;=
( x yu v
!2 GL2(K) j 2xu = 2vy = 0; uy + xv = 1
)so that
O(V; bq) = SL2(K) =
( x yu v
!2 GL2(K) j det
x yu v
!= 1
)
if char K = 2. On the other hand
O(V; q) =
( x yu v
!2 GL2(K) j xu = vy = 0; uy + xv = 1
):
Thus O(V; q) is strictly contained in O(V; bq) in characteristic 2.
Let (V1; q1); (V2; q2) be quadratic forms. We de�ne the orthogonal sum
(V; q) = (V1 ? V2; q1 ? q2)
as V = V1 � V2 and q(x) = q1(x1) + q2(x2) for x = (x1; x2) 2 V; xi 2 Vi: We
have obviously
disc(q1 ? q2) = disc(q1) � disc(q2):
1. Quadratic Forms 7
In particular q1 ? q2 is nonsingular if, and only if, q1; q2 are nonsingular. Let (V; q)
be a quadratic form (not necessarily nonsingular) and let U � V be a subspace. We
denote the restriction of q to U by qjU and we set
U? = fx 2 V j bq(x; y) = 0; 8y 2 Ug:
Lemma 3. Let (V; q) be a quadratic form and let U � V be a subspace such that
(U; qjU) is nonsingular. Then
(V; q) = (U; qjU) ? (U?; qjU?):
Proof. It clearly suÆces to show that V = U � U?. Let h : V ! V � be the
adjoint of q and let hjU : U ! U� be its restriction to U . By hypothesis it is an
isomorphism. Let now v 2 V . The restriction of h(v) 2 V � to U� is in the image of
hjU since hjU is an isomorphism. Thus there exists u 2 U such that b(x; u) = b(x; v)
for all x 2 U . This means that v� u is an element of U? or that V = U +U?. The
fact that U \ U? = f0g follows from the fact that qjU is nonsingular.
Important examples of quadratic spaces are hyperbolic spaces. The hyperbolic
planeH(K) was de�ned in Example 2. A quadratic space isometric to the orthogonal
sum of n hyperbolic planes is called a hyperbolic space of dimension 2n and is denoted
by H(Kn). The space H(Kn) is given by the class [(001n0 )], where 1n is the n� n{
identity matrix. There is another, more intrinsic, way to de�ne hyperbolic spaces.
Let V be a vector space of dimension n over K. We de�ne a quadratic form on
H(V ) = V � V �
by
q(x; f) = f(x); x 2 V; f 2 V �:
Choosing a basis fe1; : : : ; eng of V and taking the dual basis in V �, it is easy to see
that H(V ) is isometric to the hyperbolic space H(Kn). Any K{linear isomorphism
' : V ~!V 0 induces an isometry
H(') = ('; '��1) : H(V ) ~!H(V 0):
Further, we have canonical isometries H(V �W ) ' H(V ) ? H(W ).
8 1. Quadratic Forms
Let (V; q) be a quadratic space. Then (V;�q) is also a quadratic space. We have
Proposition 4. (V; q) ? (V;�q) ' H(V ).
Proof. If (V; q) is given by the class [�], then q ? �q is given by the class [(�00��)].
We have to �nd an invertible (2n� 2n){matrix � such that��t � 00 ��
!��=�
0 1n0 0
!�:
Such a matrix is given by
� =
1 �1� �t
!�1
(use that ( 0��
�t
0 ) is alternating and that [�] = [�t]).
Corollary 5. Any quadratic space is an orthogonal summand in a hyperbolic space.
Let (V; q) be a quadratic space. An element e 6= 0 2 V is called isotropic if
q(e) = 0 and a subspace U of V is called totally isotropic if qjU is identically zero.
For example the subspace V of the hyperbolic space H(V ) is totally isotropic. Con-
versely, we have
Proposition 6. Let U be a totally isotropic subspace of a quadratic space (V; q).
There exists a morphism ' : H(U) ! (V; q) whose restriction to U is the identity.
In particular (V; q) contains an orthogonal summand isometric to H(U).
Proof. The embedding U ! V induces a surjective map V � ! U�. Composing
with the isomorphism hq : V ~!V � we obtain a surjective map t : V ! U� given by
v 7! bq(v;�)jU . Its kernel is U?, i.e. the sequence
0! U? ! Vt! U� ! 0
is exact. Let W be a direct summand of U? in V such that tjW is an isomorphism
W ~!U�. It follows that the dual map U ! W �, which is given by u 7! bq(u;�)jW is
also an isomorphism. Since U � U?; U \W = f0g and U;W generate a subspace
of V which is a direct sum U �W . The restriction of bq to U �W is given by a bloc
1. Quadratic Forms 9
matrix
� =
0 �t1�1 Æ
!
if we choose a basis fe1; : : : ; e2ng of U �W such that fe1; : : : ; eng is a basis of U
and fen+1; : : : ; e2ng is a basis of W . Since � is of the form � + �t; Æ is of the form
� + �t and we can choose for � the matrix
� =
0 0�1 �
!:
If we make a change of basis in U �W given by a bloc matrix � = (10x1), we get
�0 =
0 0�1 �1x + �
!
for the new matrix. This does not change the basis of U . Since �1 is invertible, we
may choose x = ���11 �. Therefore we can assume that � = 0 or � = (0�1
00). We now
replace � by (00�t10 ), which di�ers from � by the alternating matrix ( 0
��1
�t10 ). We
recall that �t1 is the matrix of tjW : W ~!U�. Thus (100
(�t1)�1) de�nes an isomorphism
' : U � U� ~!U �W � V . Since 1 00 ��1
1
! 0 �t10 0
! 1 00 (�t1)
�1
!=
0 10 0
!;
' is a morphism H(U)! V of quadratic spaces and 'jU is the identity.
Corollary 7. Let (V; q) be a quadratic space of dimension 2m. If U � V is a totally
isotropic subspace of dimension m, then (V; q) ' H(U).
Corollary 8. Let (V; q) be a quadratic space and let e 2 V be an isotropic element.
There exists an isotropic element f 2 V such that bq(e; f) = 1. In particular V
contains a hyperbolic plane as an orthogonal summand.
A pair fe; fg of elements of V is called a hyperbolic pair if e; f are isotropic
and bq(e; f) = 1.
Let (V; q) be a quadratic form. We say that x 6= 0 2 V is anisotropic if q(x) 6= 0
and that (V; q) is anisotropic if all x 6= 0 2 V are anisotropic.
10 1. Quadratic Forms
Corollary 9. Let (V; q) be a quadratic space. There exists a decomposition
(V; q) = (V0; q0) ? H(Kn)
with (V0; q0) anisotropic.
The decomposition is unique up to isometry, in view of the following Witt can-
cellation theorem and we call n the Witt index of (V; q).
Theorem 10 (Witt cancellation).. Let (V1; q1); (V2; q2) and (V 0; q0) be quadratic
spaces such that (V1; q1) ? (V 0; q0) ' (V2; q2) ? (V 0; q0). Then (V1; q1) ' (V2; q2).
Before proving Theorem 10, we remark that Witt cancellation does not hold
for bilinear spaces in characteristic 2. We have
� =
0B@ 1 0 00 0 10 1 0
1CA =
0B@ 1 1 01 0 11 1 1
1CAt0B@ 1 1 0
1 0 11 1 1
1CAso that 13 and � de�ne isometric bilinear spaces over IF2. But (
10
01) and (01
10) are
not isometric.
Proof of Theorem 10. By Corollary 5 and induction, we may assume that we
have two orthogonal decompositions
(V1; q1) ? H1 = (V; q) = (V2; q2) ? H2
of a quadratic space (V; q) with H1; H2 hyperbolic planes. Let fe1; f1g be a hyper-
bolic pair in H1 and fe2; f2g a hyperbolic pair in H2. In view of the next propo-
sition there is an isometry ' of (V; q) such that '(e1) = e2; '(f1) = f2. Since
(V1; q1) = H?1 and (V2; q2) = H?
2 ; ' is an isometry (V1; q1) ~!(V2; q2).
Proposition 11. Let (V; q) be a quadratic space and fe1; f1g; fe2; f2g be hyper-
bolic pairs in V . There exists an isometry ' of (V; q) such that '(e1) = e2; '(f1) =
f2.
We give some de�nitions before proving Proposition 11. We say that two hy-
perbolic planes H1; H2 in (V; q) are adjacent if there exists a hyperbolic pair fe1; f1g
1. Quadratic Forms 11
in H1 and a hyperbolic pair fe2; f2g in H2 with a common element. We say that H1
and H2 are related if there is a �nite chain of adjacent hyperbolic planes connecting
H1 and H2.
Lemma 12. Two hyperbolic planes H1; H2 in a quadratic space (V; q) are always
related.
Proof. Let fe1; f1g; fe2; f2g be hyperbolic pairs in H1; H2. If one of the 4
elements bq(e1; e2); bq(e1; f2); bq(f1; e2); bq(f1; f2) is not zero, say bq(e1; e2), we
replace the pair fe2; f2g of H2 by the pair fe3; f3g; e3 = e2bq(e1; e2)�1; f3 =
f2bq(e1; e2). Then H1; H2 are adjacent to the hyperbolic plane generated by the
hyperbolic pair fe1; e3g. If all of the above 4 elements are equal to zero, then
fe1; f1+e2g; ff2; f1+e2g are hyperbolic pairs generating hyperbolic planes H3 and
H4. The chain H1; H3; H4; H2 shows that H1 and H2 are related.
Proof of Proposition 11. In view of Lemma 12, we may assume that e1 = e2.
Let f = f2 � f1 and let U = Kf . If bq(f; f1) 6= 0, then V = Kf1 � U?. The
map ' : V ! V given by '(�f1 + w) = �(f1 + f2) + w is an isometry such that
'(f1) = f2 and '(e1) = e1. If bq(f; f1) = 0, then q(f) = 0 and f 2 H?1 . Let
v 2 H?1 such that ff; vg is a hyperbolic pair in H?
1 and let H3 be the hyperbolic
plane generated by ff; vg. The linear map u1 : H1 ? H3 ! H1 ? H3 given by
e1 7! e1; f1 7! f1 + f; f 7! f and v 7! v � e1 has the matrix
� =
0BBB@1 0 0 00 1 1 00 0 1 0�1 0 0 1
1CCCAwith respect to the basis fv; f; f1; e1g. Since
�t
0BBB@0 1 0 00 0 0 00 0 0 10 0 0 0
1CCCA � =0BBB@
0 1 1 00 0 0 0�1 0 0 10 0 0 0
1CCCA �0BBB@
0 1 0 00 0 0 00 0 0 10 0 0 0
1CCCA ;we see that u1 is an isometry of H1 ? H3 such that u1(e1) = e1 and u1(f1) = f2.
Since
(V; q) ' H1 ? H3 ? (H1 ? H3)?;
12 1. Quadratic Forms
u1 can be extended to an isometry u = u1 ? 1 with the wanted property.
If the characteristic of K is not equal to 2, there is a much simpler proof of
the Witt cancellation theorem. We need the notion of a diagonalizable form. Let
V = Kn with the canonical basis fe1; : : : ; eng and let a1; : : : ; an be elements of K.
We denote by ha1; : : : ; ani the quadratic form q : V ! K given by
q(ei) = ai; i = 1; : : : ; n; bq(ei; ej) = 0; i 6= j:
A quadratic form (V 0; q0) is called diagonalizable if (V 0; q0) ' ha1; : : : ; ani for some
a1; : : : ; an 2 K. Obviously
ha1; : : : ; ani ' ha1i ? � � � ? hani:
Lemma 13. If char K 6= 2, any quadratic form is diagonalizable.
Proof. This is clear if q(v) = 0 for all v 2 V . Assume that q(v1) = a1 6= 0 and
let U = Kv1. Since char K 6= 2; (U; qjU) is nonsingular and the claim follows from
Lemma 3 and induction on the dimension.
Thus to prove Witt cancellation for quadratic spaces over a �eld of characteristic
not equal to 2, it suÆces to show that
(V1; q1) ? hai ' (V2; q2) ? hai ) (V1; q1) ' (V2; q2):
This, in turn, is a consequence of
Proposition 14. Let (V; q) be a quadratic space over a �eld K of characteristic not
equal to 2 and let x; y be elements of V such that q(x) = q(y) 6= 0. There exists an
isometry ' of V such that '(x) = y.
To prove Proposition 14 we need the notion of a re ection. Let u be an
anisotropic element in a quadratic space (V; q) (over any �eld K). The map �u :
V ! V de�ned by
�u(x) = x� bq(x; u)
q(u)u
1. Quadratic Forms 13
is called a re ection. Obviously �u(u) = �u and �u = 1 on (Ku)?. We let it as an
exercise to check that �u is an isometry of (V; q) and that � 2u = 1.
Proof of Proposition 14. If q(x) = q(y) we have the equality
bq(x + y; x+ y) + bq(x� y; x� y) = 4bq(x; x):
So either x� y or x + y is anisotropic. If x� y is anisotropic, then �x�y(x) = y. If
x+ y is anisotropic, then �x+y�x(x) = y.
Corollary 15. O(V; q) is generated by re ections if char K 6= 2.
Proof. Since (V; q) is a quadratic space over a �eld K with char K 6= 2, there exists
x 2 V anisotropic. Let ' 2 O(V; q) and let y = '(x). By Proposition 14, there
exists a product of re ections � such that �(y) = x, so (�')(x) = x. Let U = (Kx)?.
We have V = U ? Kx, since x is anisotropic (and char K 6= 2 !) and �'jU is an
isometry of U . By induction on the dimension, �'jU is a product of re ections in
U . These re ections can be extended to re ections on V (by taking the identity on
x). This proves the claim.
Remark 16. Corollary 15 is a weak form of the theorem of Cartan{Dieudonn�e.
The corollary holds also in characteristic 2, except for (V; q) = H(IF2) ? H(IF2),
where the switch of the 2 hyperbolic planes is not a product of re ections. The proof
in the general case is quite complicated (see Dieudonn�e, Sur les groupes classiques).
If the characteristic ofK is not equal to 2, any quadratic space is the orthogonal
sum of 1{dimensional spaces. In general, we have the weaker (and best possible) re-
sult that a quadratic space of even dimension is an orthogonal sum of 2{dimensional
subspaces. For a; b 2 K, let [a; b] denote the quadratic form (K2; q) with
q(e1) = a; q(e2) = b and bq(e1; e2) = 1;
where fe1; e2g is the canonical basis of K2. The form [a; b] is nonsingular if, and
only if, 1� 4ab 6= 0.
14 1. Quadratic Forms
Lemma 17. Any quadratic space (V; q) of even dimension 2m is isometric to an
orthogonal sum of m spaces [ai; bi] with 1� 4aibi 6= 0; i = 1; : : : ; m.
Proof. Assume �rst that char K = 2. Since (V; q) is nonsingular, there exist
x; y 2 V such that bq(x; y) = � 6= 0. Replacing y by 1�y, we have bq(x; y) = 1:
Let a1 = q(x); b1 = q(y). Since d(x; y) = 4a1b1 � 1 = 1, we see that: (1) x; y
are linearly independent; (2) the restriction of q to Kx � Ky is nonsingular. The
claim then follows by Lemma 3 and induction on the dimension. If char K 6= 2, it
suÆces to show that a 2{dimensional diagonal space ha; bi; a � b 6= 0, is isometric to
a space [a1; b1] with 1 � 4a1b1 6= 0. Let x1; x2 2 V with q(x1) = a; q(x2) = b and
bq(x1; x2) = 0. Let x3 = x2 +12ax1. We get
bq(x1; x3) = 1; q(x3) = b + 14a
and 1� 4a(b+ 14a) = �4ab 6= 0
so that ha; bi ' [a; b + 14a].
In characteristic 2 a quadratic space cannot have odd dimension, since the
matrix of its polar form is alternating and an (n � n){alternating matrix has zero
determinant if n is odd. To overcome this diÆculty (and since there exist interest-
ing forms in odd dimensions also in characteristic 2 !), we introduce the notion of
12{regular forms (see the notes of Kneser). We �rst de�ne the 1
2{discriminant. For
this we need the following
Lemma 18. Let n be an odd integer and let ai; 1 � i � n; and aij; 1 � i < j � n,
be indeterminates. There exists a polynomial p(ai; aij) 2 ZZ[ai; aij] such that
det
0BBBB@2a1 a12 � � � a1na12 2a2 � � � a2n...
......
a1n a2n � � � 2an
1CCCCA = 2p(ai; aij):
Proof. Since n is odd the determinant on the left hand side is zero modulo 2.
For example, we have
det
0B@ 2a1 a12 a13a12 2a2 a23a13 a23 2a3
1CA = 2(4a1a2a3 + a12a23a13 � a1a223 � a2a
213 � a3a
212):
1. Quadratic Forms 15
Let now (V; q) be a quadratic form over K with DimKV = n odd. Let
fe1; : : : ; eng be a basis of V and let ai = q(ei); aij = bq(ei; ej); i < j. We
call the element p(ai; aij), where p is as in Lemma 18, the 12{discriminant of q with
respect to the basis fe1; : : : ; eng and denote it by d0(e1; : : : en). As for the discrimi-
nant, the class of d0(e1; : : : en) modulo K�2 is independent of the choice of the basis
and we denote it by 12disc(q). If (V1; q1) is of odd dimension and (V2; q2) of even
dimension, we have
12disc(q1 ? q1) =
12disc(q1) � disc(q2):
We say that (V; q) is 12{regular if 1
2disc(q) 6= 0. A 1{dimensional form hai is 1
2{
regular if a 6= 0 and q1 ? q2 is12{regular if q1 is
12{regular of odd dimension and q2
is nonsingular of even dimension.
We have the following decomposition theorem for 12{regular forms:
Lemma 19. Let (V; q) be 12{regular of dimension 2m + 1. There exist elements
a0; a1; b1; : : : ; am; bm 2 K with a0 and 1� 4aibi; i = 1; : : : ; m, not zero, such that
(V; q) ' ha0i ? [a1; b1] ? � � � ? [am; bm]:
Proof. If char K 6= 2, the claim follows from Lemma 17 and Lemma 13. Assume
that char K = 2 and that DimKV � 3. Since (V; q) is 12{regular, bq is not identi-
cally zero and there exists x1; y1 2 V with bq(x1; y1) = 1. Then x1; y1 are linearly
independent and q restricted to Kx1 �Ky1 is nonsingular. The claim now follows
by Lemma 3 and induction.
We observe that the notion of 12{regular quadratic forms is also important in
the theory of forms over commutative rings R such that 1262 R but not necessarily
2 = 0 in R. A recent application can be found in the paper of Swan mentioned in
the bibliography.
Remark 20. The decomposition (V; q) = ha0i ? hV 0; q0i with (V 0; q0) nonsingular,
given by Lemma 19, is in fact unique in characteristic 2. The 1{dimensional space
ha0i is the radical of q, i.e.ha0i = fx 2 V j bq(x; y) = 0; 8y 2 V g
16 1. Quadratic Forms
and (V 0; q0) is the nonsingular part, i.e. V 0 is generated by all x 2 V with bq(x; z) 6= 0
for some z 2 V .
Chapter 2
Central Simple Algebras
This chapter presents basic results about �nite dimensional algebras. References
are the books of Cohn, Herstein, Jacobson (see the bibliography) or any advanced
algebra textbook. A reference for this chapter as well as for the next two chapters
is the book of Scharlau. Many proofs in this chapter are copied from Scharlau's book.
Let R be a ring. A left or right R{module M is called simple if M 6= 0 and
M has no submodules other than M and 0. The ring R is called simple if it has
no two{sided ideals other than 0 and R. By Schur's lemma the endomorphism ring
A = EndR(M) of a simple module is a skew �eld (i.e. a ring such that any non-zero
element has an inverse). We de�ne the centre Z(R) of a ring as
Z(R) = fx 2 R j ax = xa; 8a 2 Rg:
Obviously R is a Z(R){algebra. For any ring R we denote the ring of all (n � n){
matrices with entries in R by Mn(R). We de�ne the opposite ring Rop of R as
Rop = R as additive group while the product in Rop is de�ned as ao � bo = (ba)o,
where ao is a viewed as an element of Rop.
Lemma 1. Let D be a skew �eld with centre K. The ring Mn(D) is simple. Its
centre is K � 1n, where 1n is the (n� n){unit matrix.
Proof. Let Eij be the matrix with entry 1 in place (i; j) and entries zero otherwise.
Then EijEk` = ÆjkEi`, where Æjk is the symbol of Kronecker. If � is a matrix with a
non-zero entry aij, then a�1ij Eki�Ejk = Ekk. Therefore the two{sided ideal generated
by � contains the unit matrix 1n = E11+� � �+Enn. This proves thatMn(D) is simple.
18 2. Central Simple Algebras
If � = (aij) 2 Z(Mn(D)), the equation Eij� = �Eij implies that aii = ajj for all i
and j and aij = 0 for all i 6= j. Thus � 2 K �1n. This shows that Z(Mn(D)) = K �1n.
We have the following converse:
Lemma 2. Let K be a �eld and A a �nite dimensional simple K{algebra. There
exists a skew �eld D such that A 'Mn(D).
Proof. Let M be a non-zero minimal left ideal of A. Such exist since A is a �nite
dimensional vector space over K. Then M is a simple A{module and EndA(M)
is a skew �eld by Schur's lemma. Let D = EndA(M)op. We view M as a right
D{module putting m � d = d0(m). For any a 2 A, let `a : M ! M be given by left
multiplication with a. Since `a 2 EndD(M), we have a ring homomorphism
i : A! EndD(M); a 7! `a;
which is injective, since A simple. We check that i is surjective. Let x; y 2 M .
The map �y : x 7! xy is an element of D, thus f(xy) = f(x)y for x; y 2 M
and f 2 EndD(M). We claim that i(M) is a left ideal in EndD(M). We have
f � i(x)(y) = f(xy) = f(x)y = i(f(x))(y), hence f � i(x) = i(f(x)) 2 i(M) for
f 2 EndD(M) as claimed. Since A is simple, we have A =MA; so i(A) = i(M)i(A.
Therefore i(A) is a left ideal of EndD(M). Since 1 2 i(A); i is surjective. Finally,
we have EndD(M) 'Mn(D) for some n since M is a �nite dimensional vector space
over D.
Lemma 3. Let D be a skew �eld and A =Mn(D). Then
1) The ideals Li of column matrices:
Li =
0BBBB@0 � � � � 0...
......
0 � � � � 0i� th column
1CCCCA ; i = 1; 2; : : : n
are minimal left ideals of A and Mn(D) ' L1 � � � � � Ln.
2) All simple A{modules, in particular all minimal ideals of A, are isomorphic.
3) Every A{module M 6= 0 which is �nite dimensional as a D{vector space, is
2. Central Simple Algebras 19
a direct sum of simple A{modules.
Proof. 1) Let � =Pj Ejiaj 2 Li and ak 6= 0, then (a�1
k Ekk)� = Eki 2 Li. Hence
the left ideal generated by any non zero element � of Li is Li, so Li is minimal.
2) Obviously all Li are isomorphic as left A{modules. Let N be a simple A{
module. Since A =Pi Li and AN = N , there is i such that LiN 6= 0. The map
Li ! N; � 7! �x, is not the zero map for some x 2 N . Since Li and N are simple
this map is an isomorphism.
3) Let N be a maximal submodule of M such that N � M and N 6= M and
let x 2MnN . Since M=N is simple, M = N +Ax = N +Pi Lix. Here Lix is either
0 or a simple module. There is i such that Lix�
6= N . Since Lix is simple and N
is a maximal submodule of M , M = N � Lix. The claim follows by induction on
DimDM .
Lemma 4. If D; D0 are skew �elds such that Mn(D) 'Mn0(D0), then D ' D0 and
n = n0.
Proof. To show that D ' D0 it suÆces to show that for any simpleMn(D){module
N , there exists an isomorphism Dop ' EndMn(D)(N). Since two simple Mn(D){
modules are isomorphic, we can take for N the column Dn. Since Dn is also a right
D{module, we get a map
' : Dop ! EndMn(D)(Dn); '(d) = rd;
where rd is right multiplication with d. Obviously, ' is a ring homomorphism. Since
D is a skew �eld, ' is injective. We let it as an exercise to check that ' is surjective.
The next result follows from Lemma 2 and Lemma 4.
Theorem 5 (Wedderburn). Let A be a �nite dimensional simple K{algebra.
Then A 'Mn(D), for a suitable skew �eld D. The integer n is uniquely determined
by A and D is determined up to an isomorphism.
Let A be a K{algebra. For any subset B � A, we de�ne the centralizer of B
20 2. Central Simple Algebras
in A to be
ZA(B) = fa 2 A j ab = ba; 8b 2 Bg:
In particular ZA(A) is the centre of A. Obviously K � ZA(A) = Z(A) and we say
that A is central if K = Z(A). We shall use the abbreviation \c:s: algebra" for a
�nite dimensional central simple K{algebra.
Lemma 6. 1) If A; B are �nite dimensional K{algebras and A0 � A; B0 � B
are subalgebras, then ZAB(A0 B0) = ZA(A
0) ZB(B0). In particular we have
Z(A B) = Z(A) Z(B).
2) If A is a c:s: algebra and B is a simple algebra, then A B is simple.
3) A B is a c:s: algebra if and only if A and B are c:s: algebras.
Proof. 1) The inclusion � is obvious. Let fb1; : : : ; bng be a basis of B and let
x =nXi=1
xi bi 2 ZAB(A0 B0):
We have
(ax1) b1 + � � � = (x1a) b1 + � � � for all a 2 A0
hence xi 2 ZA(A0). Let fc1; : : : ; ckg be a basis of ZA(A0). Then x =Pk
i=1 ci yi; yi 2 B. The same argument shows that yi 2 ZB(B0).
2) Let I 6= 0 be a two{sided ideal of A B. We have to show that I =
A B. Assume �rst that I contains an element a b 6= 0. We have AaA = A,
hence there exist ai; a0i 2 A; 1 � i � t, such that 1 =
Pti=1 aiaa
0i. ThereforePt
i=1(ai 1)(a b)(a0i 1) = 1 b. The same argument applied to b shows that
1 1 2 I. Let now x =Pki=1 ai bi; ai 2 A; bi 2 B be an element of I with k
minimal. Suppose that k > 1. By the above argument we can assume that ak = 1.
By the minimality of k; ak�1 and ak are linearly independent. Since A is central
ak�1 62 Z(A). Let a 2 A such that aak�1 � ak�1a 6= 0 and let y be the element
(a 1)x� x(a 1) = (aa1 � a1a) b1 + � � �+ (aak�1 � ak�1a) bk�1:
The bi are linearly independent by the minimality of k and the last summand is not
zero, hence y 2 I is not zero and has a smaller k. Therefore k = 1 and we are in
the case considered above.
2. Central Simple Algebras 21
3) One direction is an immediate consequence of 1) and 2) and the other is ob-
vious: if A is not central, AB is not central. If A is not simple, AB is not simple.
An algebra A is c:s: if and only if the opposite algebra Aop is c:s. We have
Lemma 7. Let A be a c:s: algebra. The homomorphism of algebras
' : A Aop ! EndK(A); '(a bo)(x) = axb
is an isomorphism.
Lemma 7 is an immediate consequence of Lemma 6 and of the following (obvi-
ous) result
Lemma 8. Let ' : A ! B be a homomorphism of �nite dimensional K{algebras.
If DimKA = DimKB and A is simple, then ' is an isomorphism.
Let A;B be c:s: algebras over K. We say that A and B are Brauer equivalent
if there exist integers k and ` such that
AMk(K) ' B M`(K):
Let A =Mr(D) and B =Ms(D0). Then A and B are Brauer equivalent if, and only
if, D ' D0. Let [A] be the equivalence class of A. We get a well-de�ned multiplica-
tion by putting [A] � [B] = [AB]. The multiplication is commutative, associative,
has [K] = [Mn(K)] as element 1 and each element has an inverse, since [A][Aop] = 1.
Thus the equivalence classes of c:s: algebras form a group. It is called the Brauer
group of K and is denoted by Br(K).
Example 9. We have Br(K) = f1g for any algebraically closed �eld. In fact, let
D be a central division algebra over K, i.e. a skew �eld D with centre Z(D) = K
and which is �nite{dimensional over K. If K � D and K 6= D, let d 2 DnK. We
get a �eld extension K � K(d) � D which is �nite over K, so d is algebraic over K
and d 2 K. Contradiction.
If K = IR, a classical result of Frobenius says that IR; IC and the quaternions IH
22 2. Central Simple Algebras
are the only �nite dimensional division algebras over IR. Since IC is commutative, any
c:s: algebra over IR is of the formMn(D) withD = IR or IH. Hence Br(IR) = ZZ=2ZZ.
If A is a K{algebra and a is a unit of A, i.e. an invertible element, then
ia(x) = axa�1; x 2 A;
is an automorphism of A. Such automorphisms are called inner. We now prove a
very useful classical result.
Theorem 10 (Skolem-Noether).. Let A be a c:s: algebra over K and B a �nite
dimensional simple K{algebra. Let �; : B ! A be two algebra homomorphisms.
There exists an inner automorphism ia of A such that = ia Æ �.
Proof. Let �rst A = EndK(V ); V a vector space over K. We can view V as a left
B{module in two di�erent ways, putting b �v = �(b)(v) or b �v = (b)(v). By Lemma
3 these two B{modules are isomorphic so there exists a K{linear isomorphism f :
V ! V such that
f(�(b)(v)) = (b)f(v) for all v 2 V
hence (b) = f�(b)f�1. For the general case, we consider the maps
� 1; 1 : B Aop ! A Aop = EndK(A):
The algebra B Aop is simple by Lemma 6. It then follows from the case A =
EndK(V ) that there exists f 2 A Aop such that
(b) ao = f(�(b) ao)f�1
for all b 2 B; ao 2 Aop. Setting b = 1, we see that f 2 ZAAop(1 Aop) = A 1
(by Lemma 6). Therefore we get f = a 1. The claim �nally is a consequence of
(b) 1 = (a 1)(�(b) 1)(a 1)�1.
Corollary 11. Let A be a c:s: algebra. Every homomorphism A ! A is an inner
automorphism.
2. Central Simple Algebras 23
Remark 12. Let A be a c:s: algebra over K and let S be a �nite dimensional
commutative K{algebra. We claim that S{automorphisms of B = S A are inner.
This generalization of Corollary 11 will be used later. Let Rad(S) be the Jacobson
radical of S i.e. the set of nilpotent elements of S. The quotient S=Rad(S) is a
product of �eld extensions of K. Let now ' be an automorphism of B and let ' be
the induced automorphism of B = (S=Rad(S)) A. By an obvious generalization
of Corollary 11, ' is inner, ' = iv; v 2 B�. Let
J' = fx 2 B j '(b)x = xb; 8b 2 Bg:
The automorphism ' is inner if, and only if, there exists u 2 J' which is a unit of
B. We get
J' = (S=Rad(S)) J' = J'
= fx 2 B j '(b)x = xb; 8b 2 Bg
and
J' = Sv since ' = iv:
Let v 2 B be such that its image in B is v. By Nakayama's Lemma (see any of
the algebra textbooks in the bibliography or Scharlau's book), v is a unit of B and
J' = Sv. This shows the claim.
Corollary 13. Let A; B; C be c:s: algebras. Then A B ' A C implies that
B ' C.
Proof. Let E = A C and let � be an isomorphism AB ~!A C. We have
C = fx 2 E j x(a 1) = (a 1)x; 8a 2 Ag = ZE(A 1)
�(B) = fy 2 E j y�(a 1) = �(a 1)y; 8a 2 Ag = ZE(�(A 1)):
Let now : A ! E be given by (a) = a 1 and ' : A ! E by '(a) = �(a 1).
By Theorem 10 there exists u 2 E such that = iu Æ '. It follows that iu induces
an isomorphism �(B) ~!C.
Lemma 14. 1) Let K � L be a �eld extension and let A be a K{algebra. Then
A is a c:s: K{algebra if and only if L A is a c:s: L{algebra. In particular K � L
24 2. Central Simple Algebras
induces a group homomorphism Br(K)! Br(L).
2) Let A be a c:s: K{algebra. There exists a �nite �eld extension K � L such
that L A 'Mn(L). In particular DimKA is always a square.
Proof. 1) follows from Lemma 6 (at least one direction of 1) !).
2) Let K be the algebraic closure of K. By Example 9, we get that K A 'Mn(K). Let � be such an isomorphism. Let fx1; : : : xrg be a basis of A over K,
then
�(1 x`) =Xi;j
�ij`Eij; �ij` 2 K;
where fEij j 1 � i; j � ng, is the canonical basis of Mn(K). Let L be the sub�eld
of K generated by K and the �ij`. Since the �ij` are algebraic over K; L is a �nite
extension of K and � can be viewed as an isomorphism L A ~!Mn(L).
We call a pair (L; �); � : LA ~!Mn(L), a splitting of A. The following result
is a nontrivial sharpening of Lemma 14. We refer to any of the books mentioned at
the beginning of this chapter for a proof (for example Herstein or Scharlau).
Theorem 15. For any c:s: K{algebra A, there exists a splitting (L; �) with K � L
a �nite Galois extension.
We now use the existence of Galois splittings to construct the characteristic
polynomial of a c:s: algebra. Let (L; �) be a Galois splitting of A. If L[X] is the
polynomial ring in one variable X over L, we extend K{automorphisms of L to
K[X]-automorphisms of L[X] by putting g(X) = X for all g 2 G = Gal(L=K). As
for �elds, we have
K[X] = fp(X) 2 L[X] j g(p(X)) = p(X); 8g 2 Gg:
Further, we can extend the isomorphism � : L A ~!Mn(L) to an isomorphism
� : L[X] A ~!Mn(L[X])
by putting �(X 1) = X � 1n. For a 2 A, we de�ne the reduced characteristic
polynomial of a by
�(X; a) = det(�(X 1� 1 a)):
2. Central Simple Algebras 25
Let �(X; a) = Xn + �n�1Xn�1 + � � � + �0. The coeÆcient (�1)n�0 is called the
reduced norm of a. We denote it by n(a) or nA(a). The coeÆcient { �n�1 is the
reduced trace and it is denoted by tr(a) or trA(a).
Lemma 16. The polynomial �(X; a) is independent of the choice of the Galois
splitting (L; �) and has coeÆcients in K. In particular n(a) 2 K and tr(a) 2 K for
a 2 A.
Proof. Let � : L A ~!Mn(L) and �0 : L0 A ~!Mn(L
0) be two Galois splittings.
Replacing L and L0 by a Galois extension L00 which contains L and L0, we see that
we may assume L = L0. In view of Corollary 11, we have �0 = iu Æ � for some
u 2 GLn(L) so that
det(�0(X 1� 1 a)) = det(u(�(X 1� 1 a))u�1)
= det(�(X 1� 1 a)):
We �nally show that �(X; a) 2 K[X]. For any g 2 Gal(L=K), let g be the K{
automorphism of Mn(L[X]) given by
g(aij) = (g(aij)):
The automorphism �(g 1)��1g�1 of Mn(L[X]) is L{linear, hence there is u 2GLn(L) such that �(g 1)��1g�1 = iu. Then we get
�(1 a) = ug(�(1 a))u�1:
Since det(g(aij)) = g(det(aij)), it follows that �(X; a) = g(�(X; a)) for all g 2 G.
Therefore �(X; a) 2 K[X], as claimed.
The reduced norm and trace satisfy the following formulas
n(ab) = n(a)n(b) tr(a+ b) = tr(a) + tr(b)n(�a) = �nn(a) tr(�a) = �tr(a)
tr(ab) = tr(ba)
for a; b 2 A and � 2 K. Further a 2 A is invertible if and only if n(a) 6= 0. These
properties follow from the corresponding properties of the determinant and the trace
for matrices. Further �(a; a) = 0 for all a 2 A, i.e. the Cayley{Hamilton theorem
holds for the characteristic polynomial.
Chapter 3
Involutions on Central Simple Algebras
Let R be a ring. As already mentioned in Chapter 1, Remark 1, an involution
on R is a map � : x 7! x of R to itself such that
1) a+ b = a+ b 2) ab = ba; 3) a = a:
for all a; b 2 R. A homomorphism ' : (R; �) ! (R0; �0) of rings-with-involution is
a homomorphism ' : R! R0 such that ' Æ � = �0 Æ '.
Let now A be aK{algebra,K a �eld. If � is an involution of A, the restriction of
� to K is an involution �0 of K. If �0 is the identity, we say that � is an involution of
the �rst kind. Hence involutions of the �rst kind are K{linear. If �0 is not the iden-
tity, letK0 be the �xed �eld of �0,K0 = fx 2 K j �0(x) = xg: The extensionK0 � K
is a quadratic Galois extension with Galois group Gal(K=K0) = f1; �0g ' ZZ=2ZZ.
If this case occurs we say that � is an involution of the second kind.
We now study involutions of the �rst kind on c:s: algebras and begin with
the algebra A = Mn(K). For any matrix x, the map x 7! xt is an involution of
Mn(K) of the �rst kind. If � is another such involution, the map x 7! �(xt) is
an automorphism of Mn(K). Therefore, by the Skolem{Noether theorem, there is
u 2 GLn(K) such that
�(xt) = uxu�1 or �(x) = uxtu�1; x 2Mn(K):
The fact that �2 = 1 implies ut = "u, where " = �1. We denote the involution
x 7! uxtu�1 by �u and call " the type of �u. This is well de�ned, since �u = �v
implies u = �v for 0 6= � 2 K. Involutions of type 1 are usually called of orthogonal
type and involutions of type { 1 of symplectic type. If char K = 2 there are no
3. Involutions on Central Simple Algebras 27
di�erences between the two types. In this case it is useful to consider a more special
class of involutions. We say that �u is of even "�type if u is of the fom v + "vt. In
particular �u is of even symplectic type if u is an alternating matrix
u =
0BBBB@0 u12 : : : u1n
�u12 0...
. . .�u1n 0
1CCCCA ;not just a skew{symmetric matrix. We observe that involutions of even symplectic
type can only occur if n is even. This is clear if charK 6= 2. If charK = 2, an
alternating matrix is the matrix of the polar of a quadratic form. Its determinant is
a multiple of 2 if n is odd (Lemma 18 of Chapter 1) hence is zero in characteristic
2. Thus an (n� n){alternating matrix, n odd, is never invertible. Let
Altn(K) = fy 2Mn(K) j y is alternating g = fx� xt j x 2Mn(K)g:
The vector space Altn(K) is of dimension n(n�1)2
over K. Let now � = �u be an
involution of Mn(K) of type ". We call
Alt�n(K) = fx� "�(x) j x 2Mn(K)g:
the space of alternating elements of �. We have for y 2 Alt�n(K),
y = x� "�(x) = x� "uxtu�1 = (xu� "uxt)u�1
= (xu� (xu)t)u�1
and similarly y = u(u�1x� (u�1x)t). Thus
Lemma 1. We have Alt�n(K) = Altn(K) �u�1 = u �Altn(K) if � = �u. In particular
DimKAlt�n(K) = n(n�1)
2:
Let �u; �v be two involutions on Mn(K) and let
' : (Mn(K); �u) ~!(Mn(K); �v)
be an isomorphism of K{algebras{with{involution. By Corollary 11 of Chapter 2,
' = ic; c 2 GLn(K), and c�u(x)c�1 = �v(cxc
�1) for x 2 Mn(K):
A computation shows that v�1cuct is in the centre of Mn(K). Hence there exists
� 2 K� such that
�v = cuct:
28 3. Involutions on Central Simple Algebras
In particular the type of the involution �u is an invariant of the isomorphism class
of (Mn(K); �u). Another consequence is the following
Lemma 2. Let sm be the 2m{alternating matrix with m blocs ( 0�1
10) on the diag-
onal and let v be any invertible alternating matrix. There exists c 2 GL2m(K) such
that v = csmct. In particular (M2m(K); �v) ' (M2m(K); �sm).
Proof. Let V = K2m and let b : V � V ! K be the alternating bilinear form
de�ned by b(�; �) = �tv�; � = (x1; : : : ; x2m)t and � = (y1; : : : ; y2m)
t. Since v is
invertible, there exist �1; �2 2 V such that b(�1; �2) 6= 0 and we can even assume
that b(�1; �2) = 1. The two vectors �1 and �2 are linearly independent. Let now
V1 = K�1 �K�2 and V2 = f� 2 V j b(�i; �) = 0; i = 1; 2g:
We have V = V1 � V2. Let f�3; : : : ; �2mg be a basis of V2 and let d = (dij) be the
matrix of the change of basis, i.e., �j =Pdijei, where fe1; : : : ; e2mg is the canonical
basis of K2m. Putting v0 = (b(�i; �j)), we have
v0 =
s1 00 v00
!and v0 = dtvd:
The matrix v00 is alternating and the claim then follows by induction on the dimen-
sion of V .
We now discuss involutions of the �rst kind on c:s: algebras. Let A be such
an algebra, with involution � : a 7! a, and let (L; �) be a splitting of A. Then
�(1�)��1 is an involution of Mn(L), hence of the form �u; u 2 GLn(L). Let " bethe type of �u. Using that " depends only on the isomorphism class of (Mn(L); �u)
it is easy to check that " is independent of the choice of the splitting (L; �). We call
" the type of the involution �. Again, " depends only on the isomorphism class of
(A; �). In the next Lemma we collect some results whose proofs are straightforward.
Lemma 3. Let A1; A2 be K{algebras with involutions �1( resp. �2).
1) Then �1 �2 is an involution of A1 A2 of the �rst kind if �1; �2 are of the
�rst kind and of second kind if �1; �2 are of the second kind.
2) If �1 has type "1 and �2 type "2, then �1 �2 has type "1"2.
3. Involutions on Central Simple Algebras 29
3) If u is a unit of A1 such that �1(u) = �u; � = �1 and if �1 is of type "1,
then iu Æ �1 is an involution of type "1�. Conversely if iu Æ �1 is an involution for
some unit u of A1, then �1(u) = �u.
Let A be a c:s: algebra of dimension n2 with an involution � of type ". We call
the set
Alt�(A) = fx� " �(x) j x 2 Ag
the set of alternating elements of A. Let � : L A ~!Mn(L) be a splitting of A
such that �(1 �)��1 = �u. By Lemma 1, we have �(Alt�(A)) = u � Altn(L), thusAlt�(A) is a vector space of dimension n(n�1)
2over K.
We say that � is of even "{type if there is a splitting (L; �) such that �(1 �)��1 = �u with u = v + "vt 2 GLn(L). This de�nition is in fact independent
of the splitting. If (L0; �0) is another splitting, we may assume that L0 = L and
�0 = ic �, for some c 2 GLn(L). If �0(1�)�0�1 = �u0, with u0 2 GLn(L), we have
ic Æ �u Æ i�1c = �u0 , hence �u
0 = cuct for some unit � of L. We have �u0 = v0 + "v0t,
where v0 = cvct. If " = �1, we say that the involution is of even symplectic type
and if " = 1, we say that the involution is of even orthogonal type. In fact, if the
involution is of even symplectic type, by Lemma 2, we can even choose the splitting
(L; �) such that �(1 �)��1 = �sm with sm = diag(( 0�1
10); : : : ; (
0�1
10)), where
4m2 = DimKA. We further observe that 1 2 Alt�(A) if � is of even symplectic type.
Going over to a splitting of A as above, it suÆces to remark that s�1m �1 2 Alt2m(K).
In the following examples, we now discuss two important classes of K{algebras
with K{linear involutions, the quadratic algebras and the central simple algebras of
dimension 4.
Example 4. Quadratic Algebras. We say that a K{algebra S is quadratic if
DimKS = 2. Let f1; vg be a basis of S. We have
v2 = av + b
for some a; b 2 K. Hence
S ' K[X]=(X2 � aX � b):
30 3. Involutions on Central Simple Algebras
It follows, in particular, that a quadratic algebra is always commutative. We de�ne
a K{linear map � : x 7! x of S by
�v + � = �v + � and v = a� v:
It is easy to check that v2 = v2, so � is an involution of S. we have
�v + �+ �v + � = �a+ 2�;
and
(�v + �)(�v + �) = �2vv + ��(v + v) + �2
= �b�2 + a��+ �2:
Therefore x + x 2 K and xx 2 K for all x 2 S. By the following Lemma 6, there
exists exactly one involution � of S with
x + �(x) 2 K and x�(x) 2 K for all x 2 S:
We call it the standard involution of S. We denote x+ x by tr(x) or trS(x) and xx
by n(x) or nS(x). The map n : S ! K is a quadratic form and its polar form
bn(x; y) = n(x + y)� n(x)� n(y)
has the matrix 2 aa �2b
!with respect to the basis f1; vg of S. We say that S is a separable quadratic K{
algebra if n is nonsingular. If K � L is a �eld extension, then S is separable
quadratic over K if and only if L S is separable quadratic over L. Since
�det
2 aa �2b
!= a2 + 4b
is the discriminant of the polynomial X2 � aX � b, S is a separable quadratic K{
algebra if and only if S ' K[X]=p(X) for some separable quadratic polynomial
p(X). Let now S be separable quadratic. Since
n(xy) = xyxy = xyy x = n(x)n(y);
S is a �eld if and only if n is anisotropic. If n is isotropic, it follows from Corollary
8 of Chapter 1 that the quadratic space (S; n) is isometric to H(K). Thus there
3. Involutions on Central Simple Algebras 31
exists e; f 2 S such that ee = 0; ff = 0 and ef + fe = 1. It follows from
x+ x = tr(x) 2 K and xx = n(x) 2 K that
x2 � tr(x)x + n(x) = 0; x 2 S:
Let now 1 = �e+�f; �; � 2 K. We have also 1 = �e+�f , hence e = �fe; f = �ef
and ef = 0. It follows that e = �e2. Since e2 � tr(e)e + n(e) = 0, we get e+ e = 1�
and �f = �e. Similarly, we have f = �f 2. Now 1 = ef +fe = �
�f 2+ �
�e2 = 1
�f + 1
�e,
so �� = 1; � = e+ �2f = e+ f = � and �2 = �2 = 1. Finally we get
1 = �2e2 + �2f 2 = e2 + f 2; e2f 2 = 0
and fe2; f 2g is a pair of orthogonal idempotents in S. Therefore
S ' K �K with the involution (�; �) 7�! (�; �):
To summarize, we have shown that a separable quadratic K{algebra S is either a
quadratic �eld extension of K or is isomorphic to K � K. The �rst case occurs if
the norm of S is anisotropic and the second if the norm is isotropic. Finally we
observe that for any separable quadratic K{algebra S; S S ' S � S. An explicit
isomorphism is given by x y 7! (xy; xy); x; y 2 S.
Example 5. Quaternion Algebras. We de�ne a quaternion algebra over a �eld
K to be a c:s: K{algebra A of dimension 4. In view of Theorem 5 of Chapter 2,
A is either a central division algebra over K or A ' M2(K). The reduced norm
is a quadratic form on A, the trace is a linear form on A and the characteristic
polynomial reduces to
�(X; a) = X2 � tr(a)X + n(a):
If A =M2(K), the reduced norm is the determinant. Its polar bdet has the matrix0BBB@0 �1 0 0�1 0 0 00 0 0 10 0 1 0
1CCCAwith respect to the basis e1 = E12; e2 = E21; e3 = E11 and e4 = E22. Therefore det is
a nonsingular quadratic form on M2(K). The subspace U = (KK00) of M2(K) is to-
tally isotropic. Thus we get by Corollary 7 of Chapter 1 (or by a direct computation
32 3. Involutions on Central Simple Algebras
!) an isometry of quadratic spaces
(M2(K); det) ' H(U) ' H(K) ? H(K) = H(K2):
For any matrix � = (acbd) 2M2(K), let � = tr(�)� �. We have
� =
d �b�c a
!=
0 1�1 0
! a bc d
!t 0 1�1 0
!�1
:
Thus � 7! � is an involution of even symplectic type of M2(K) such that
� + � = tr(�) 2 K and �� = �� = det(�) 2 K
for all � 2M2(K). By the following Lemma 6, � 7! � is the unique involution � of
M2(K) with the property that � + �(�) 2 K and ��(�) 2 K for all � 2 M2(K).
Let now A be any c:s: K{algebra of dimension 4 and let � : A! A be de�ned by
�(a) = a = trA(a)� a:
Let � : LA ~!M2(L) be a splitting of A. Since trA(a) is the trace of the matrix �(1a), the map �(1 �)��1 is the involution de�ned above for A =M2(K). Therefore
� is an involution of even symplectic type. We call � the standard involution of A.
We have nA(a) = a�(a) = �(a)a since it holds over a splitting and � is the unique
involution of A such that
a+ �(a) 2 K and a�(a) 2 K for all a 2 A:
Let ' : A ~!B be an isomorphism of c:s: algebras of dimension 4 and let �0 = '�A'�1
where �A is the standard involution of A. The map �0 is an involution of B such that
x+ �0(x) 2 K and x�0(x) 2 K for all x 2 B. Hence �0 = �B and ' is automatically
an isomorphism (A; �A) ~!(B; �B) of algebras{with{involutions. It follows that '
induces an isometry (A; nA) ~!(B; nB) of the reduced norms. In particular, a splitting
of A, � : L A ~!M2(L) induces an isometry
L (A; nA) = (L A; nLA) ~!(M2(L); det):
Since (M2(L); det) is nonsingular over L; (A; nA) is nonsingular. We have nA(xy) =
nA(x)nA(y) for x; y 2 A and nA(1) = 1. Thus A is a division algebra if and only if
nA is anisotropic. Since either A is a division algebra or A = M2(K), we see that
nA is isotropic if, and only if, A =M2(K).
3. Involutions on Central Simple Algebras 33
Let A be a K{algebra. We say that a K{linear involution � on A is a standard
involution if
a+ �(a) 2 K and a�(a) 2 K for all a 2 A:
(In fact it suÆces to assume that a�(a) 2 K for all a 2 A, since (a + 1)�(a + 1) =
a�(a) + a + �(a) + 1 if A has an element 1).
Lemma 6. A K{algebra A can carry at most one standard involution.
Proof. Let a�(a) = n(a) and a+ �(a) = tr(a). We have
a2 � tr(a)a+ n(a) = 0
in A for any standard involution �. Let now �1; �2 be standard involutions on A
and let tri(a) = a + �i(a); ni(a) = a�i(a). Let fe1; : : : ; eng be a basis of A with
e1 = 1. Since
e2i � tr1(ei)ei + n1(ei)e1 = e2i � tr2(ei)ei + n2(ei)e1
we get tr1(ei) = tr2(ei) for i � 2, hence �1(ei) = �2(ei) for i � 2. On the other
hand, �1(1) = 1 = �2(1), thus, as claimed �1 = �2.
The existence of standard involutions is related with the existence of com-
position algebras. Let A be a �nite dimensional K{vector space with a bilinear
multiplication (a; b) 7! ab which has a neutral element 1, i.e. a � 1 = a = 1 � a. In
the following, we call A an algebra even if the multiplication is not associative. We
say that A is a composition algebra if there exists a nonsingular quadratic form n
on A such that n(ab) = n(a)n(b) for all a; b 2 A. An associative K{algebra with
1, which carries a standard involution �, is a composition algebra for the quadratic
form n(a) = a�(a), since
n(ab) = ab�(ab) = ab�(b)�(a) = a�(a)b�(b) = n(a)n(b):
The following result, which is due to Hurwitz in characteristic not 2, describes all
possible composition algebras. It implies that separable quadratic algebras and c:s:
algebras of dimension 4 are the only examples of associative algebras with a stan-
dard involution such that the corresponding norm is nonsingular.
34 3. Involutions on Central Simple Algebras
Theorem 7. Let (A; n) be a composition algebra over K. Then either
1) A = K,
2) A is a separable quadratic K{algebra and n = nA,
3) A is a c:s: K{algebra of dimension 4 and n = nA,
4) A is a separable Cayley algebra. In particular DimKA = 8.
Proof. We give the proof of van der Blij and Springer, which works also in charac-
teristic 2. Let (x; y) be the polar of n, i.e. (x; y) = n(x+ y)� n(x)� n(y). Since byassumption n is nonsingular,
(x; y) = 0; 8y 2 A ) x = 0:
The following formulas are deduced from n(xy) = n(x)n(y) by linearization:
(xy1; xy2) = n(x)(y1; y2)
(x1y; x2y) = n(y)(x1; x2)(3.1)
(x1y2; x2y1) + (x1y1; x2y2) = (x1; x2)(y1; y2):
We must have n(1) = 1. Putting x1 = y2 = x; x2 = z and y1 = 1 in the third
formula of (3.1), we obtain
(x2 � (1; x)x+ n(x) � 1; z) = 0
for all z 2 A. Since n is nonsingular we get
x2 � (1; x)x + n(x) � 1 = 0(3.2)
for all x 2 A. We call (1; x) the trace of x and denote it by tr(x). Let x = tr(x)�x.It is straightforward to check that
xy = y x; x = x and 1 = 1:
Hence x 7! x is an involution on A. Further
xx = xx = n(x); n(x) = n(x); (xy; z) = (y; xz) = (x; zy):(3.3)
We also need the formulas
x(xy) = n(x)y and (yx)x = n(x)y:(3.4)
3. Involutions on Central Simple Algebras 35
For the proof of the �rst one, we have
(x(xy); z) = (xy; xz) = (n(x)y; z) for all z 2 A:
The proof of the other one is similar. It follows from (3.4) that x(xy) = (xx)y.
Therefore
x(xy) = x(tr(x)y � xy) = (xtr(x)� xx)y = (xx)y
and similarly (yx)x = y(xx). This shows that A is an alternative algebra. Assume
now that DimK(A) > 1. Since n is nonsingular, it is easy to check that there exists
u 2 A such that the set f1; ug is linearly independent in A and such that the re-
striction of n to B1 = K1�Ku is nonsingular. By (3.2), B1 is a separable quadratic
K{algebra. We have A = B1 ? B?1 . If B
?1 6= 0, the restriction of n to B?
1 is non-
singular and there exists v 2 B?1 such that n(v) 6= 0. We put n(v) = ��; � 2 K.
To conclude we use the following:
Lemma 8. Let B be a proper subalgebra of A such that the restriction of n to B
is nonsingular and let v 2 B? be such that n(v) = �� is not zero. Then B � vB is
a subalgebra of A. We have
n(a+ vb) = n(a)� �n(b) and a + vb = a� vb
for a; b 2 B. Further B is associative and B is commutative if A is associative.
Proof. The fact that B � vB is a subalgebra follows from the formulas
(va)b = v(ba)
a(vb) = v(ab)(3.5)
(va)(vb) = �ba n(v) = �ba
for a; b 2 B. We only check the �rst one. The proof of the others is similar. We
have v = �v, since (v; 1) = 0, and 0 = (v; a) = va + av = �va + av, thus va = av
for a 2 B. Further
((vb)a; z) = (vb; za) = (bv; za) = �(ba; zv):
The last equality follows from the third formula of (3.3), using that (v; a) = 0 for
a 2 B. On the other hand
�(ba; zv) = �((ba)v; z) = (v(ab); z):
36 3. Involutions on Central Simple Algebras
This holds for all z 2 A, hence (vb)a = v(ab) as claimed. The formulas for n and
the involution are easy. Let now
n((a+ vb)(c+ vd)) = n(ac + �db+ v(cb+ ad))
= n(ac + �db)� �n(cb+ ad):
On the other hand
n((a+ vb)(c+ vd)) = n(a + vb)n(c+ vd)
= (n(a)� �n(b))(n(c)� �n(d)):
Comparing both expressions and using once more that n is multiplicative, we obtain
(ac; �db) + n(v)(cb; ad) = 0
or
(ac; db) = (ad; cb)
then it follows from (3.3) that
((ac)b; d) = (a(cb); d) for all a; b; c; d 2 B:
Thus (ac)b = a(cb) and B is associative. If A is associative, we have (va)b = v(ab) =
v(ba) and ba = ab. Therefore B is commutative.
We now go back to the proof of Theorem 7. The algebra B2 = B1 � vB1 is an
associative K{algebra of dimension 4. We let it as an exercise to check that B2 is
central simple. If B2 is a proper subalgebra of A, let B3 = B2 + wB2 with w 2 B?2 .
If B3 = A, the formulas (3.5) show (by de�nition!) that A is a Cayley algebra (see
the book of Schafer). If B3 is a proper subalgebra of A, we get B4 = B3 + zB3 with
z 2 B?3 . By Lemma 8, B3 must be associative and hence B2 commutative. This is
not possible.
Let A be a c:s: K{algebra with an involution � of the �rst kind. Since � is
an isomorphism A ~!Aop, the class of A in Br(K) is of order � 2. Conversely any
c:s: K{algebra A such that its class in Br(K) is of order � 2 carries an involution
� of the �rst kind. More precisely:
3. Involutions on Central Simple Algebras 37
Theorem 9 (Albert). Let A be a c:s: K{algebra such that 2[A] = 0 in Br(K):
1) There exists an involution � of the �rst kind on A: 2) If the dimension of A is
even, the involution � can be chosen of even orthogonal type or of even symplectic
type.
Proof. We follow the proof given by Saltman. If 2[A] = 0 in Br(K); [A] = [Aop]
so there exists an isomorphism � : A ~!Aop of K{algebras. Let
' : A A ~!EndK(A)
be the isomorphism de�ned by '(a b)(x) = ax�(b). Let now !A be the switch of
A A, i.e. !A(a b) = b a. By the Skolem{Noether theorem, !A = iu for some
u 2 (AA)�. If A = EndK(V ); AA = EndK(V V ) and u : V V ! V V can
be chosen as the switch !V . If � : LA ~!EndL(V ) is a Galois splitting, the element
u = (� �)�1(!V ) 2 L A A is in fact in A A (this can easily be checked by
Galois theory). Thus u 2 (A A)� can be chosen such that (� �)(u) = !V for
any Galois splitting. In particular u2 = 1. Let now : A ! A be given by '(u).
We have
2 = 1 and (ax�(b)) = b (x)�(a) for a; b; x 2 A:
Lemma 10. Let w = (1) 2 A: 1) We have 1 = w�(w) = �(w)w and � 2 = (iw)�1.
2) If there exists a unit a 2 A such that (a) = a, then � = ia Æ � is a K{linear
involution of A.
Proof. 1) We have 1 = (w) = (w � 1) = w�(w). On the other hand 1 =
(1 �w) = ��1(w)w. Thus w is a unit of A with inverse �(w) = ��1(w). Further we
have x = 2(x) = ( (x � 1)) = (w�(x)) = (�(x) � 1) � �(w) =w� 2(x)�(w) = (iw Æ � 2)(x): 2) We have �(xy) = �(y)�(x) since � is an antiau-
tomorphism. We check that �2 = 1. It follows from a = (a) = (1 � a) = �(a)w
that w = a�(a)�1. Thus
�2 = (ia Æ �)2 = ia�(a)�1 Æ � 2 = 1 since � 2 = iw�1:
Proof of Theorem 9. To prove the �rst part, we have, by Lemma 10, to construct
a unit a 2 A such that (a) = a. We may assume that A = Mn(D); D a division
algebra. Since 2[D] = 0 in Br(K) there exists an antiautomorphism � : D ~!D. If
38 3. Involutions on Central Simple Algebras
� is an involution, we extend � to A by taking the transpose on Mn(K). If � is
not an involution, we construct as above a map : D ! D such that 2 = 1 and
(ax�(b)) = b (x)�(a) for a; b; x 2 D. Since �2 6= 1; �2 = i�1w for w = (1)
implies that w 62 K. Thus 1 + (1) 6= 0 and a = 1 + (1) is the wanted element a.
If A 'Mn(K); A has an involution given by the transpose.
We now prove the second claim of Theorem 9. By 1) A has an involution �,
say of type ". Let � = �1. If A contains a unit of the form u = z + "��(z), then
�0 = iuÆ� is of type � and in particular, �0 will be of even symplectic type if � = �1.Thus to prove 2) it suÆces to show that for some K{linear involution � of A and
any " = �1, there exists a unit of A of the form u = z + "�(z). Assume that
A = Mn(D); D a division algebra, D 6= K. By 1) D has an involution �1 and it
obviously suÆces to �nd z1 2 D such that z1 + "�1(z1) 6= 0. Since D 6= K; �1 6= 1
so z1 + "�1(z1) = 0 for all z1 2 D is impossible. If A = Mn(K) with n = 2m we
write A =M2(K)Mm(K), take � = �1 �2, where �1 is the standard involution
of M2(K) and �2 is transposition. Then for z = (1000) 1; z + "�(z) = (10
0") 1
is a unit of A.
One can also give conditions for the existence of involutions of the second kind
on c:s: algebras. For this we need the notion of corestriction. Let Z be a separable
quadratic K{algebra with standard involution �0 and let A be a K{algebra. Then
Z A is a Z{algebra and, by Galois theory, A can be identi�ed with the set of
elements y 2 ZA such that (�01)(y) = y. The map ~� = �01 is a �0{semilinear
automorphism of Z A such that ~�2 = 1. Conversely, let B be a Z{algebra with a
�0{semilinear automorphism ~� such that ~�2 = 1. Let
A = fy 2 Bj~�(y) = yg:
Then A is a K{algebra and the multiplication map � : ZA! B; ya 7! ya is an
isomorphism of Z{algebras such that ~� = �(�01)��1. We say that A is descended
from B (and ~�) by Galois descent. Obviously Galois descent also applies to other
structures, like quadratic forms. If B is a Z{algebra, we denote by �0B the algebra
with the Z{action twisted through �0, i.e. y � b = �0(y)b; y 2 Z; b 2 B. The map
~� : B Z �0B ! B Z �0B
3. Involutions on Central Simple Algebras 39
given by ~�(b c) = c b is �0{semilinear and the corresponding K{algebra
A = fx 2 B Z �0B j ~�(x) = xg
is called the corestriction of B and is denoted by cor(B). If B has a Z{linear in-
volution �, then cor(B) has an induced K{linear involution cor(�). We can now
formulate a well-known existence criterion for involutions of the second kind:
Theorem 11 (Albert, Riehm).. Let L be a separable quadratic extension of K
and let A be c:s: over L. Then A admits an involution of the second kind if and
only if [cor(A)] = 0 in Br(K).
Theorem 11 will not really be used in the following and we do not prove it. A
proof can be found in the book of Scharlau (p. 309).
Let K � L be a separable quadratic extension and let A be a c:s: L{algebra
with a �xed involution x 7! x� of the second kind. In the next Lemma we describe
all possible other involutions of the second kind on A.
Lemma 12. Let � be an involution of the second kind on A.
1) There exists g 2 A�, with g� = g, such that �(x) = gx�g�1.
2) Let g 2 A� such that g� = g and let �g(x) = gx�g�1. Then �g is an involution
of the second kind and �g = �g0 if and only if g = �g0 for � 2 K�.
Let ' : (A; �g) ~!(A; �g0) be an isomorphism of L{algebras-with-involution.
There exist c 2 A� and � 2 K� such that g0 = �cgc�.
Proof. 1) By Skolem{Noether, there exists g 2 A� such that �(x�) = ig(x). It then
follows from �2 = 1 that g� = �g for some � 2 L such that ��0(�) = 1. By the
following Lemma (which is a special case of Hilbert Theorem 90) there exists � 2 Lsuch that � = ��1�0(�). Replacing g by ��1g, we see that we can assume g� = g.
The claims 2) and 3) are similar to the corresponding claims for involutions of the
�rst kind and we do not check them.
Lemma 13. Let � 2 L be such that ��0(�) = 1. There exists � 2 L such that
40 3. Involutions on Central Simple Algebras
� = ��1�0(�).
Proof. If there exists � 2 L such that � = �0(�)+�0(�)� 6= 0, then �0(�)��1 = �. If
not, � 7! ��0(�)� is a K{linear automorphism of L. This can only occur if � = �1(put � = 1!). Let then z be a generator of L such that �0(z) = 1� z. The element
� = 1� 2z is such that �0(�) = �� and � = �1 = ��1�0(�).
Chapter 4
The Cli�ord Algebra
Let (V; q) be a quadratic form over a �eld K. We would like to �nd a K{algebra
C and a K{linear map i : V ! C such that
i(x)2 = q(x) � 1C for all x 2 V:(4.1)
We de�ne the Cli�ord algebra C(V; q) of (V; q) as the K{algebra which is universal
with respect to the property (4.1): A Cli�ord algebra for (V; q) is a K{algebra C to-
gether with a K{linear map i : V ! C verifying (4.1), such that for any K{algebra
A and any K{linear map ' : V ! A with '(x)2 = q(x) � 1A, there exists a unique
K{algebra homomorphism '0 : C ! A such that ' = '0 Æ i.
Lemma 1. For any quadratic form (V; q) over K, there exists up to isomorphism a
unique Cli�ord algebra (C(V; q); i).
Proof. The uniqueness follows from the universal property of the Cli�ord algebra
and we let its proof as an exercise. We prove existence. Let
T (V ) = K � V � V V � � � �
be the tensor algebra of V , let J be the 2{sided ideal of T (V ) generated by all
elements x x� q(x) � 1 for x 2 V and let C = T (V )=J . We claim that C together
with the canonical map
i : V ! TV ! TV=J
is a Cli�ord algebra. Let ' : V ! A be a K{linear map such that '(x)2 = q(x) � 1A.By the universal property of the tensor algebra, ' extends to a unique homomor-
phism ~' : TV ! A of K{algebras such that ~'jV = '. We have ~'(J ) = 0, so ~'
42 4. The Cli�ord Algebra
induces, as claimed, a unique homomorphism '0 : C ! A such that ' = '0 Æ i.
Since a Cli�ord algebra is uniquely determined up to a unique isomorphism we
shall speak in the following of \the" Cli�ord algebra of (V; q) and we shall denote it
by C(V; q) or C(q). From now on we identify K � 1C with K.
We observe that (4.1) implies
i(x)i(y) + i(y)i(x) = bq(x; y); x; y 2 V(4.2)
where bq is the polar of q.
We say that a K{algebra A is ZZ=2ZZ{graded (or simply graded) if A has a direct
sum decomposition A = A0�A1, as K{vector space, such that Ai �Aj � Ai+j; i+ j
mod 2. We have K � 1 � A0. Such a grading on the tensor algebra TV is given by
(TV )0 = K � V V � � � �(TV )1 = V � V V V � � � �
i.e. TV is graded by the degree. The generators x x � q(x) of J have even
degree. Therefore we get a ZZ=2ZZ{grading of C(q), C(q) = C0 � C1; where C0
is generated by the even products of elements of V and C1 by the odd products.
In particular C0 is a subalgebra of C, sometimes called the even Cli�ord algebra,
and i(V ) � C1. By de�nition a graded homomorphism � : A0 � A1 ! A00 � A0
1
of ZZ=2ZZ{graded algebras is an algebra homomorphism such that �(Ai) � A0i. An
immediate consequence of the universal property of the Cli�ord algebra is that any
morphism ' : (V; q)! (V 0; q0) of quadratic forms induces a graded homomorphism
C(') : C(q) ! C(q0). In particular any isometry ' : (V; q) ~!(V 0; q0) induces an
isomorphism C(') : C(q) ~!C(q0). As we shall see later the converse does not hold,
i.e. non isometric forms may have isomorphic Cli�ord algebras.
Lemma 2. Let K � L be a �eld extension. There is a canonical isomorphism
C(L (V; q)) ' L C(V; q) for any quadratic form (V; q).
Proof. The map 1 i : LV ! LC(V; q) induces a homorphism C(L(V; q))!L C(V; q). The homomorphism V ! L V ! C(L (V; q)) yields the inverse
4. The Cli�ord Algebra 43
homomorphism.
We now compute the Cli�ord algebra of an orthogonal sum. For this we need
the graded tensor product A bB of two ZZ=2ZZ{graded K{algebras A = A0�A1 and
B = B0 � B1. As a K{vector space A bB = A B. The product is de�ned for
homogeneous elements a; a0 2 A; b; b0 2 B by
(a b)(a0 b0) = (�1)@(b)@(a0)aa0 bb0;
where @(b) is the degree (0 or 1) of b, and is extended by linearity to arbitrary
elements. The algebra A bB is graded by
(A bB)0 = A0 B0 + A1 B1
(A bB)1 = A1 B0 + A0 B1:
Let A = A0 � A1 be a graded algebra. We call the grading
M2(A)0 =
A0 A1
A1 A0
!and M2(A)1 =
A1 A0
A0 A1
!
of M2(A) the checker-board grading. On M2(K) the checker-board grading is
M2(K)0 =
K 00 K
!and M2(K)1 =
0 KK 0
!:
Let B = B0 � B1 be any ZZ=2ZZ{graded algebra. As an exercise in the theory of
graded algebras (and because it will be used later) we prove the following
Lemma 3. There exists an isomorphism of graded algebras
' :M2(K) bB ~!M2(K)B =M2(B):
Proof. We de�ne '(x 1B) = x 1B for x 2M2(K) and
'(1 (b0 + b1)) =
b0 + b1 00 b0 � b1
!2 M2(B):
Proposition 4. C(q ? q0) ' C(q) bC(q0) as graded algebras.
Proof. Let y = (x; x0) 2 V � V 0. The map y 7! i(x) 1 + 1 i(x0) induces a map
C(q ? q0) ! C(q) bC(q0). On the other hand, the inclusions x 7! (x; 0) 2 V � V 0
and x0 7! (0; x0) 2 V �V 0 extend to algebra homomorphisms C(q)! C(q ? q0) and
44 4. The Cli�ord Algebra
C(q0) ! C(q ? q0). These two maps combine to a homomorphism C(q) bC(q0) !C(q ? q0) which is the inverse of the �rst map.
Let fe1; : : : ; eng be a basis of V . In view of the relations (4.1) and (4.2) the 2n
elements 1; i(e1); : : : ; i(en); i(e1)i(e2); : : : ; i(en�1)i(en); i(e1)i(e2)i(e3); : : : ;
i(e1) � � � i(en) form a set of generators of C(V; q) as a vector space over K. A cele-
brated theorem says that they form a basis of C(V; q).
Theorem 5 (Poincar�e{Birkho�{Witt).. If fe1; : : : ; eng is a basis of V , the set
f1; i(ej1) � � � i(ejr); 1 � r � n; 1 � j1 < j2 � � � < jr � ng is a basis of C(V; q).
Proof. If the theorem is true for q1 and q2 it is true for q1 ? q2 by Proposition 4. The
theorem is true if (V; q) = hai because, if v is a generator of V with q(v) = a, then
TV ' K[X]; J ' (X2 � q(v)) and C(q) ' K[X]=(X2 � q(v)) = K � 1�K � i(v):Hence the theorem is true if fe1; : : : ; eng is an orthogonal basis. If the theorem is
true for a given basis, it is clearly true for any basis of V . By Lemma 13 of Chapter
1 the theorem is true for any �eld of characteristic not 2. Let R be a domain of char-
acteristic not 2. By embedding R into its �eld of fractions, we see that the theorem
is true for R. If now K is a �eld of characteristic 2, we can write K as a quotient
R=I with R a domain of characteristic zero (for example a polynomial ring over ZZ).
Let V be the free module with basis a set of symbols fe1; : : : ; eng, so V = V =IV .
We de�ne a quadratic form q on V by choosing elements ai; aij (i 6= j) in R such
that ai + I = q(ei) and aij + I = bq(ei; ej). We have (V ; q)R R=I ' (V; q) and by
Lemma 2 (or better an obvious extension of Lemma 2 for forms over commutative
rings) C(V ; q)R R=I ' C(M; q). The theorem is true for C(V ; q) because R is a
domain of characteristic zero. Therefore it is also true for (V; q).
Remark 6. This proof of the Poincar�e{Birkho�{Witt theorem, which is due to
Kneser, works obviously also for quadratic forms over �nitely generated free R{
modules, R any commutative ring. In fact we have used the notion of quadratic
forms over rings in the proof as well as some basic facts, which are straightforward
generalizations of the corresponding facts for forms over �elds.
4. The Cli�ord Algebra 45
Corollary 7. If V has dimension n, then C(V; q) has dimension 2n and C0; C1 have
both dimension 2n�1.
It follows from Theorem 5 that i : V ! C(V; q) is injective. From now on
we shall identify V with its image i(V ) in C(q). We now describe the structure
of the Cli�ord algebras for nonsingular even dimensional forms and 12{regular odd
dimensional forms. We put C = C(V; q); C0 = C0(V; q) and C1 = C1(V; q).
Theorem 8. 1) Let (V; q) be a quadratic space of even dimension 2m over K. Then
C is a c:s: K{algebra, the centre Z(C0) of C0 is a separable quadratic K{algebra.
If Z(C0) is a �eld, C0 is c:s: over Z(C0). If Z(C0) ' K � K, there exists, up to
isomorphism, a unique c:s: K{algebra A such that C0 ' A� A as K �K{algebra.
2) Let (V; q) be 12{regular of odd dimension 2m + 1. Then C0 is c:s: over K,
the centre Z(C) of C is a ZZ=2ZZ{graded quadratic K{algebra Z0 � Z1 such that
Z0 = K and Z1 is generated by an element z with z2 = � � 1; � 2 K�. Further the
multiplication in C induces an isomorphism Z(C) C0 ~!C.
Proof. We �rst prove Theorem 8 in dimension 1 and 2, then by induction for forms
(V; q) = H(K) ? (V 0; q0) and �nally for arbitrary forms.
Case n = 1 : We have (V; q) = hai; a 6= 0. Let v 2 V with q(v) = a. Then
C = K � 1 +Kv; C0 = K � 1; C1 = Kv and the multiplication is given by v2 = a.
The algebra C is commutative, so Z(C) = C.
Case n = 2 : We assume that (V; q) = [a; b]. Let V = Ku � Kv with q(u) =
a; q(v) = b and bq(u; v) = 1. Then C has the basis f1; u; v; uvg with the relations
u2 = a; v2 = b; uv + vu = 1:
To show that C is c:s:, it suÆces to show that L C is c:s: over L for some �eld
extension L of K. Let e = �u + v; � a symbol. We get q(e) = �2a + � + b and
we choose L such that the polynomial X2a + X + b has a root � in L. Then e is
isotropic in L V . By Corollary 8 of Chapter 1, L (V; q) ' [0; 0] and L C is
generated by elements u; v with u2 = 0; v2 = 0 and uv+vu = 1. It is easy to verify
46 4. The Cli�ord Algebra
that
u 7! (0010); v 7! (01
00)
induces an isomorphism
L C ~!M2(L):
Hence C is c:s: over K. We have in fact proved that
C(H(K)) ' M2(K);
as graded algebras, ifM2(K) has the checker{board grading. This will be used later.
Let now (V; q) = [a; b] as above. We have C0 = K �1�Kuv and (uv)2 = u(1�uv)v =uv � ab. Thus the element z = uv is a generator of C0 such that
z2 = z � ab:
By Example 4 of Chapter 3, C0 is a quadratic algebra and is separable if and only
if 1 � 4ab 6= 0. Since the discriminant of q with respect to the basis fu; vg is
4ab � 1; Z(C0) = C0 is a separable quadratic K{algebra. In particular Z(C0) is
either a �eld or Z(C0) ' K �K.
This shows that Theorem 8 holds in dimension 1 and 2. We assume now that
(V; q) = H(K) ? (V 0; q0). Let C 0 = C 00 � C 0
1 be the Cli�ord algebra of (V 0; q0). We
get
C 'M2(K) bC 0
by Proposition 4 and the fact that C(H(K)) ' M2(K). Thus by Lemma 3
C 'M2(K) C 0 =M2(C0):
If we use the isomorphism C ~!M2(C0) given by Lemma 3 to identify C withM2(C
0),
then
C0 =
C 00 C 0
1
C 01 C 0
0
!and C1 =
C 01 C 0
0
C 00 C 0
1
!in M2(C
0):
Assume that V has even dimension, so V 0 also has even dimension. By induction,
Theorem 8 holds for (V 0; q0). In particular C is c:s: over K since C 0 is c:s: over K.
We now construct an inner automorphism ofM2(C0) which maps C0 toM2(C
00). Let
x 2 V 0 be anisotropic (such an element exists since q0 is nonsingular). Then x is
invertible in C 0 with inverse x�1 = x�q(x)�1 2 C 01. Left and right multiplication with
4. The Cli�ord Algebra 47
x induce isomorphisms C 01 ~!C 0
0 of K{vector spaces. In particular xC 01 = C 0
0; C01x =
C 00 and xC
00x = C 0
0. Thus inner conjugation with the unit (100x) of M2(C
0) maps
C0 =
C 00 C 0
1
C 01 C 0
0
!to
C 00 C 0
1xxC 0
1 xC 00x
!=M2(C
00):
It follows that Z(C0) ' Z(C 00) and that C0 is as described in 1). If (V; q) is
12{regular
of odd dimension, then (V 0; q0) is also 12{regular of odd dimension. We have as above
C 'M2(C0) and C0 'M2(C
00)
so that, by induction on the dimension, C0 is c:s: over K and Z(C) ' Z(C 0) is a
quadratic algebra. Since the isomorphism C 'M2(C0) is an isomorphism of graded
algebras, Z(C) ' Z(C 0) as graded algebras. The last claim Z(C) C0 ~!C is also
clear by induction.
If (V; q) is arbitrary, there exists a �eld extension K � L such that L (V; q) 'H(L) ? (V 0; q0). Therefore Theorem 8 holds for L (V; q). Assume that V has
even dimension. The algebra C is c:s: over K since L C is c:s: over L and Z(C0)
is separable quadratic since Z(L C0) = L Z(C0) is separable quadratic over L.
If Z(C0) is a �eld, it also follows from Theorem 8 that C0 is c:s: over Z(C0). If
Z(C0) = K �K, we only get that there exists an isomorphism of K �K{algebras
C0 ~!A� B; A; B c:s: over K with L A ' L B. To show that A ' B, we use
the following
Lemma 9. Let (V; q) be nonsingular of even dimension. There exists an isomor-
phism
' : Z(C0) C ~!EndC0(C) =M2(C0)
of Z(C0){algebras, where C is viewed as a right C0{module through the multiplica-
tion in C.
Proof. We de�ne '(zc)(x) = cxz for z 2 Z(C0); c; x 2 C. If Z(C0) is a �eld, ' is
an isomorphism since Z(C0)C is a c:s: algebra over Z(C0) and Z(C0)C; M2(C0)
have the same dimension. A similar argument works if Z(C0) = K � K. The fact
that EndC0(C) ' M2(C0) follows from C1 = C0x for x 2 V anisotropic.
48 4. The Cli�ord Algebra
We go back to the proof of Theorem 8. If Z(C0) = K � K and C0 = A � B,
we have by the above lemma
M2(A)�M2(B) ' C � C:
Since M2(A); M2(B) and C are simple algebras, it follows that
M2(A) ' C 'M2(B);
hence A ' B by Corollary 13 of Chapter 2.
Finally if V is 12{regular of odd dimension, C0 is c:s: over K since L C0 is
c:s: over L. Further Z(C) is quadratic. Let z = z0 + z1; zi 2 Ci be a generator of
Z(C). The part z0 of degree zero lies in the centre of C0, hence is a scalar, so z1
also generates Z(C) as an algebra. Since z21 6= 0 in L Z(C); z21 6= 0 and Z(C) is
as claimed.
Chapter 5
Invariants of Quadratic Forms
In this chapter and the next the fact that we do not assume char K 6= 2 has a
strong in uence on the presentation of the results. Demazure{Gabriel is a useful ref-
erence for these two chapters. Parts of Proposition 5 are copied from Micali{Revoy.
Let (V; q) be a quadratic form which is either nonsingular of even dimension or
12{regular of odd dimension. Let
Z(V; q) = Z(q) =
(Z(C0(V; q)) if Dim V is evenZ(C(V; q)) if Dim V is odd.
We call Z(q), which is a graded quadratic algebra, the discriminant algebra of (V; q).
The algebra Z(q) is separable if Dim V is even or if Char K 6= 2. Clearly any
isometry (V; q) ~!(V 0; q0) induces a graded isomorphism Z(q) ~!Z(q0). By the proof
of Theorem 8, Chapter 4, we have
Z(H(K) ? q) ' Z(q) and Z(H(K)) ' K �K;
hence Z(H(U)) ' K �K for any hyperbolic space H(U).
Lemma 1. Z(q) = ZC(C0).
Proof. The claim follows from Theorem 8, Chapter 4 and Lemma 6, Chapter 2, if
DimV is odd. Assume that DimV is even. The inclusion � is clear. Let x 2 ZC(C0)
and let x = x0 + x1 be its decomposition in homogeneous parts. Since C0 is ho-
mogeneous, x0 and x1 must lie in ZC(C0). By Proposition 5 of this chapter, we
have a generator z of Z(q) such that x1z + zx1 = x1, so (1 � 2z)x1 = 0. But
(1� 2z)2 = 1+ 4(z2 � z) = 1+ 4r is a unit (Proposition 5), hence x1 = 0. (The use
50 5. Invariants of Quadratic Forms
of Proposition 5 in the proof is not very elegant but is legal since Lemma 1 is not
applied in the proof of Proposition 5 !).
We call the isomorphism class of Z(q) the Arf invariant of (V; q) and denote it
by a(q). We say that the Arf invariant a(q) is trivial if Z(q) ' K �K.
Example 2. Assume that char K 6= 2 and that fe1; : : : ; eng is an orthogonal basis
of (V; q). The element z = e1 � � � en is a generator of Z(q) and
z2 = (�1)n(n�1)2 a1 � � �an = (�1)n(n�1)
2 d(e1; : : : ; en):
Let Z = K � 1 + K � z and Z 0 = K � 1 + K � z0 be two quadratic algebras with
z2 = a and (z0)2 = b. We claim that Z ' Z 0 if and only if a = �2b for some � 2 K�
(assuming char K 6= 2 !). Let ' : Z ~!Z 0 be an isomorphism and let '(z) = �z0 + �.
Since ' is an isomorphism, � 6= 0. We have '(z2) = '(z)2, so a = �2b+2��z0 + �2.
Since 2� 6= 0; � = 0 and a = �2b as claimed. Therefore Z(q) is determined up to
isomorphism by the class of (�1)n(n�1)2 a1 � � �an in K�=K�2 if char K 6= 2. This class
is equal to disc(q) multiplied by the class of 2n(�1)n(n�1)2 in K�=K�2, since disc(q)
is the class of 2na1 � � �an. For any a 2 K�, let [a] denote its class in K�=K�2. The
element
Æ(q) = [(�1)n(n�1)2 ] � disc(q) 2 K�=K�2
is called the signed discriminant of (V; q). It follows from the above discussion that,
for forms of the same dimension, Z(q) ' Z(q0) if and only, if Æ(q) = Æ(q0) (if char
K 6= 2).
The following property of the discriminant algebra follows from Theorem 8 of
Chapter 4 or from Lemma 1 and the fact that C(L (V; q)) ' LC(V; q); C0(L(V; q)) = L C0(V; q):
Lemma 3. For any �eld extension K � L; Z(L (V; q)) ' L Z(V; q).
We now construct a generator z of Z(q), for any characteristic, given orthogonal
decompositions (see Chapter 1)
(V; q) = [a1; b1] ? � � � ? [am; bm] if DimV = 2m;
5. Invariants of Quadratic Forms 51
(V; q) = [a1; b1] ? � � � ? [am; bm] ? ha2m+1i if DimV = 2m+ 1:
We construct z by induction, using the following
Lemma 4. Let (V1; q1) and (V2; q2) be quadratic spaces of even dimensions, and let
Z1 = Z(q1); Z2 = Z(q2). For i = 1; 2, we assume that Zi has a generator zi such
that
z2i = zi + ri; ri 2 K with 1 + 4ri 6= 0
and
zivi + vizi = vi for all vi 2 Vi:
Then Z = Z(q1 ? q2) has a generator z such that
z2 = z + r with r = r1 + r2 + 4r1r2; hence 1 + 4r = (1 + 4r1)(1 + 4r2)
and
zv + vz = v for all v 2 V = V1 � V2:
Proof. Let z = z1 1 + 1 z2 � 2z1 z2 in C(q1 ? q2) = C(q1) bC(q2). We get
z2 = z + r with r = r1 + r2 + 4r1r2. The other claims follow by straightforward
computations.
Let (V; q) = [a; b] and let fx; yg be a basis of V such that q(x) = a; q(y) = b
and bq(x; y) = 1. The discriminant algebra Z(q) = C0 is generated by z = xy
and z2 = z � ab. It is easy to verify that zv + vz = v for all v 2 V . Further
1 � 4ab = �d(x; y) is not zero. Thus we can apply Lemma 4 to quadratic spaces
V1; V2 of type [a; b] or to orthogonal sums of such spaces. Let Vi = [ai; bi] and
(V; q) = (V1; q1) ? : : : ? (Vm; qm):
We choose in each Vi a basis fei; ei+mg such that
q(ei) = ai; q(ei+m) = bi and bq(ei; ei+m) = 1
and put zi = eiei+m; ri = �q(ei)q(ei+m) = �aibi. By Lemma 4 (and induction) the
element
z =mXj=1
(�2)j�1Sj(z1; : : : ; zm);
52 5. Invariants of Quadratic Forms
where Sj is the j � th elementary symmetric function in m letters, is a generator of
Z(q) such that
z2 = z + r; r =mXj=1
4j�1Sj(r1; : : : ; rm)
and zv + vz = v for all v 2 V . Further
1 + 4r =mYi=1
(1 + 4ri) = (�1)md(e1; e1+m; : : : ; em; e2m):
Therefore the class of 1 + 4r in K�=K�2 is the signed discriminant Æ(q) of (V; q).
Let now (V; q) be of odd dimension. We assume that
(V; q) = (V1; q1) ? � � � ? (Vm; qm) ? ha2m+1i = (V 0; q0) ? ha2m+1i
with Vi as above. Let e2m+1 2 V be a basis of ha2m+1i such that q(e2m+1) = a2m+1.
By the even dimensional case, we have a generator z0 of Z(q0) � C0(q0) such that
z02 = z0 + r0 and z0v0 + v0z0 = v0 for all v0 2 V 0. The element
z = e2m+1 (1� 2z0)
of C(ha2m+1i) bC(q0) = C(q) is of degree 1 and commutes with all elements of C0(q),
thus z 2 Z(q). Since
z2 = a2m+1(1 + 4r0) = (�1)md0(e1; : : : ; e2m; e2m+1);
the element z is a homogeneous generator of Z(q). We call the class of
(�1)md0(e1; : : : ; e2m+1) in K�=K�2 the signed 1
2{discriminant of (V; q) and denote it
by 12Æ(q). The element 1
2Æ(q) depends only on the isometry class of (V; q). By the
above discussion, we have for spaces of the same odd dimension
Z(q) ' Z(q0) (as graded algebras), 12Æ(q) = 1
2Æ(q0):
Observe that if (V; q) = [a1; b1] ? � � � ? [am; bm] ? ha2m+1i and char K = 2, then
12Æ(q) = [a2m+1] 2 K�=K�2. Summarizing, we have
Proposition 5. 1) Let (V; q) be a nonsingular quadratic form of even dimension.
There exists an element z 2 C0 such that
i) Z(C0) = K � 1�K � z; z2 = z + r; r 2 K and 1 + 4r 2 K�.
ii) The class of (1 + 4r) in K�=K�2 is the signed discriminant of q.
5. Invariants of Quadratic Forms 53
iii) vz + zv = v for all v 2 V , hence xz + zx = x for all x 2 C1.
2) Let (V; q) be 12{regular of odd dimension. There exists an element z 2 C1 such
that
i) Z(C) = K � 1�K � z; z2 = s; s 2 K�.
ii) The class of s in K�=K�2 is the signed 12{discriminant of q.
The quadratic algebra Z(q) has a unique standard involution �0. We now study
how �0 is related to the structure of the Cli�ord algebra C(q). By the universal prop-
erty of the Cli�ord algebra, there exists a unique K{linear involution � of C(q) such
that �(v) = �v for v 2 V . We call � the standard involution of C(q) (even if � is
not necessarily a standard involution in the sense of Chapter 3). The involution �0
of C which is the identity on V will be called the canonical involution of C. Fur-
ther, let C(�1) be the automorphism of C(q) such that C(�1)(v) = �v for all v 2 V .
Proposition 6. 1) If (V; q) is 12{regular of odd dimension, then �0 is the restriction
of C(�1) to Z(q): 2) Assume that (V; q) is nonsingular of even dimension. Then
i) �0(x)y = yx for all x 2 Z(q) and y 2 C1.
ii) If the dimension of V is congruent to 2 (mod 4), �0 is the restriction of
the standard involution � of C(q).
iii) If the dimension of V is congruent to 0 (mod 4), the standard involution
� of C(q) induces the identity on Z(q).
Proof. Let z 2 Z(q) be as in Proposition 5. The claims follow from Proposition
5, using that �0(z) = �z if the dimension of V is odd and �0(z) = 1 � z if the
dimension of V is even. Observe that we cannot describe �0 if DimV � 0(4).
In characteristic 2, the discriminant does not give any information on q or Z(q),
since Æ(q) = 1 for any nonsingular form. On the other hand the element r 2 K of
Proposition 5 can be used to characterize Z(q). We have
Lemma 7. Let K be a �eld of characteristic 2 and let Z = K � 1 � K � z; Z 0 =
K �1+K �z0 be quadratic K{algebras with z2 = z+r; z02 = z0+r0; r; r0 2 K. Then
Z and Z 0 are isomorphic if and only if there exists � 2 K such that r0� r = �2+�.
54 5. Invariants of Quadratic Forms
Proof. Let ' : Z ! Z 0 be an isomorphism given by '(z) = �z0+�. It follows from
'(z2) = ('(z))2 that �2 = � and �2r0�r = �2+�. Hence � = 1 and r0�r = �2+� as
claimed. Conversely we de�ne an isomorphism ' : Z ! Z 0 by putting '(z) = �z0+�.
The set of elements of K of the form �2 + �; � 2 K is an additive subgroup
of K if char K = 2. We denote it by }(K). Let (V; q) be a quadratic space and let
z 2 Z(q) be a generator as given by Proposition 5, i.e. z2 = z + r; r 2 K. The
class of r in the additive group K=}(K) is the classical Arf invariant of (V; q). We
denote it by �(q). In view of Lemma 7, �(q) is an invariant of the isometry class of
(V; q) and by Lemma 4
�(q1 ? q2) = �(q1) + �(q2):
If (V; q) = [a1; b1] ? � � � ? [am; bm], then
�(q) = a1b1 + � � �+ ambm:
In particular �(q) = 0 for any hyperbolic space (V; q).
The following formulas hold for orthogonal sums:
Lemma 8. 1) If (V2; q2) is nonsingular of even dimension and (V1; q1) is any
quadratic form, then
C(q1 ? q2) ' C((1 + 4r2)q1) C(q2);
where r2 is as in Proposition 5 (for (V2; q2)).
2) If (V2; q2) is12{regular of odd dimension and (V1; q1) is any quadratic form, then
C0(q1 ? q2) ' C(�s2q1) C0(q2);
where s2 is as in Proposition 5 (for (V2; q2)). In particular
C0(h�i ? q) ' C(��q)
for any quadratic form q : V ! K.
5. Invariants of Quadratic Forms 55
Proof. 1) Let z2 and r2 be as in Proposition 5 (for (V2; q2)). The element w2 = 1�2z2satis�es �0(w2) = �w2 and w2
2 = 1 + 4r2, where �0 is the standard involution of
Z(q2). We de�ne a homomorphism
' : C(q1 ? q2) = C(q1) bC(q2)! C((1 + 4r2)q1) C(q2)
by '(v1 1) = v1 w2 and '(1 v2) = 1 v2 for vi 2 Vi (apply Proposition 6,
the universal property of the graded tensor product and the universal property of
the Cli�ord algebra !). Since V1 and V2 are contained in the image of ', ' is an
isomorphism.
2) Let z2 2 Z(q2); s2 2 K be as in Proposition 5 (for q2). We de�ne
: C(�s2q1) C0(q2)! C0(q1 ? q2)
by (1 x) = j2(x), where ji : C(qi) ! C(q1 ? q2) is the canonical embedding,
i = 1; 2; and (v1) = j1(v1)j2(z2) for v1 2 V1. The map is surjective, hence an
isomorphism.
Another important invariant of quadratic forms is the Witt invariant w(q),
which takes values in the Brauer group Br(K) of the �eld K. We de�ne
w(q) =
([C(q)] 2 Br(K) if q is nonsingular of even dimension[C0(q)] 2 Br(K) if q is 1
2-regular of odd dimension.
The Witt invariant obviously depends only on the isometry class of q. Further, by
the proof of Theorem 8, Chapter 4, we have
w(H(K) ? q) = w(q) and w(H(K)) = 1
so w(H(U)) = 1 for any hyperbolic space H(U). We now compute the Witt in-
variant of an orthogonal sum q1 ? q2 with q2 of even dimension. For any class
Æ 2 K�=K�2 and any quadratic form q, the isometry class of Æ � q is well de�ned.
With this in mind, we get from Lemma 8:
Proposition 9. Let q2 be a nonsingular form of even dimension.
1) If q1 is nonsingular of even dimension, then
w(q1 ? q2) = w(Æ(q2)q1)w(q2):
56 5. Invariants of Quadratic Forms
2) If q1 is12{regular of odd dimension, then
w(q1 ? q2) = w(q1)w(�12Æ(q1)q2):
Let (V; q); (V 0; q0) be quadratic forms and let � 2 K�. A K{linear isomorphism
' : V ~!V 0 such that q0('(x)) = �q(x) for all x 2 V is called a similitude with mul-
tiplier �. Two similar forms have isomorphic even Cli�ord algebras:
Lemma 10. Let ' : (V; q) ~!(V 0; q0) be a similitude with multiplier �. There exists
an isomorphism C0(') : C0(q) ~!C0(q0) such that
C0(')(xy) = ��1'(x)'(y) for x; y 2 V:
Proof. Let T 0(V ) be the even part of the tensor algebra of V . The algebra isomor-
phism T 0(') : T 0(V )! T 0(V 0), de�ned by T 0(')(x0� � �x2n) = ��n'(x0) � � �'(x2n),induces the wanted isomorphism.
We describe now another useful isomorphism of Cli�ord algebras. First we need
a de�nition. Let Z be a separable quadratic K{algebra with standard involution
�0 : z 7! z and let � 2 K�. As in Lemma 8 of Chapter 3, we de�ne a K{algebra
A = Z � uZ
of dimension 4 by the multiplication rules
(z1 + uz2)(z3 + uz4) = z1z3 + �z2z4 + u(z2z3 + z1z4):
The algebra A is c:s: of dimension 4, hence a quaternion algebra over K and has
the grading A0 = Z, A1 = uZ. We denote it by (�; Z=K].
Proposition 11. Let (V; q) be nonsingular of even dimension. There exists an
isomorphism of graded algebras
M2(K) C(V; �q) ~!(�; Z(q)=K] C(V; q):
In particular w(�q) = w(q) �w((�; Z(q)=K]).
Proof. Let Z = Z(q) and let '1 : V ! (�; Z=K] C(V; q) be given by '1(v) =
uv; v 2 V: By the universal property of the Cli�ord algebra ' extends to a homo-
morphism ' : C(V; �q)! (�; Z=K]C(V; q) which is injective since C(V; �q) is c:s:
5. Invariants of Quadratic Forms 57
Let A be the commutant of '(C(V; �q)) in (�; Z=K] C(V; q), so A C(V; �q) '(�; Z=K] C(V; q): The algebra A is generated by (Z Z)�0�0 ' K �K and by
u, so A ' (K �K)� u(K�K) ' (�;K�K]. We claim that (�;K�K] 'M2(K):
In fact an isomorphism is induced by K �K ! K 00 K
!and u 7! (01
�0).
Chapter 6
Special Orthogonal Groups and Spin
Groups
Let (V; q) be a quadratic form and let O(q) be its orthogonal group. We denote
by AutgK(C) the group of automorphisms of the Cli�ord algebra C = C0�C1 of (V; q)
as a graded algebra. Any ' 2 O(q) induces an element � = C(') 2 AutgK(C) such
that �V � V . Conversely, let � 2 AutgK(C) be such that �V � V . Then ' = �jVis an automorphism of V such that q('(v)) = '(v)2 = �(v2) = q(v); v 2 V , hence
' 2 O(q). Therefore we can identify O(q) with the subgroup of elements � of
AutgK(C) such that �V � V :
O(q) = f� 2 AutgK(C) j �V � V g:
We assume in the following that (V; q) is either 12{regular of odd dimension or
nonsingular of even dimension. Let Z(q) be the discriminant algebra of (V; q), i.e.
the centre of C if the dimension of V is odd or the centre of C0 if the dimension
of V is even. For any ' 2 O(q) the restriction of ' to Z(q) is an automorphism of
Z(q) as a graded algebra. Thus the restriction de�nes a group homomorphism
� : O(q)! AutgK(Z(q)):
Lemma 1. The group AutgK(Z(q)) is either trivial (if the standard involution �0 of
Z(q) is the identity) or has two elements 1; �0.
Proof. If DimV is even, Z(q) is concentrated in degree zero and has a generator z
such that z2 = z + r for some r 2 K. Let � be a K{automorphism of Z(q) and let
�(z) = �z + �. It follows from �(z2) = �(z)2 that
1 = �+ 2� and �+ r = �2r + �2:
6. Special Orthogonal Groups and Spin Groups 59
Substituting � = 1� 2� in the second equation we get
�2(4r + 1) = �(4r + 1):
By Proposition 5 of Chapter 5, 4r + 1 6= 0, so � = 1 or � = 0 and �(z) = �z + 1 or
�(z) = z. If DimV is odd, Z(q) is generated by an element z of degree 1 such that
z2 = s; s 6= 0 2 K. Thus, if � 2 AutgK(Z(q)), we must have �(z) = �z and �2 = 1.
In view of Lemma 1, we may de�ne a homomorphism O(q) ! ZZ=2ZZ, the
so-called Dickson map, by putting:
Dick(') =
(0 if C(')jZ(q) = 11 if C(')jZ(q) 6= 1:
The kernel of the Dickson homomorphism is called the special orthogonal group and
is denoted by SO(q). This is not the usual de�nition of SO(q) as kernel of the
determinant map, which can only be used if the characteristic is not equal to 2. We
need a di�erent de�nition to include the case of characteristic 2. However the next
lemma shows that the two de�nitions are related. Any ' 2 O(q) is represented by a
matrix (also denoted ') if we �x a basis of V . The determinant of ' is independent
of the choice of the basis. Thus the determinant induces a homomorphism
det : O(q)! K�:
Lemma 2. Assume that (V; q) is either 12{regular of odd dimension or nonsingular
of even dimension. Then
1) det(') = �1 if ' 2 O(q).2) SO(q) � Ker(O(q)
det! f�1g).3) SO(q) = Ker(O(q)
det! f�1g) if (V; q) is 12{regular or char K 6= 2.
Proof. Let fe1; : : : ; eng be a basis of V and let (uij) be the matrix of ' 2 O(q) withrespect to the basis, i.e.
'(ej) =X
uijei:
We have det(uij)2d(e1; : : : ; en) = d(e1; : : : ; en) 6= 0 if q is nonsingular and
det(uij)2d0(e1; : : : ; en) = d0(e1; : : : ; en) 6= 0 if q is 1
2{regular. Thus det(uij)
2 = 1
and the �rst claim is proved. For the second claim, we choose a basis of (V; q) as
in Lemma 17 (resp. 19) of Chapter 1 and choose a generator z of Z(q) as given by
60 6. Special Orthogonal Groups and Spin Groups
Proposition 5 of Chapter 5. Let ' 2 O(q). We claim that
C(')(z) = det(')z + �
for some � 2 K. This is clear if char K = 2, since det(') = 1 by 1) and z 7! z; z 7!z + 1 are the only possible automorphisms of Z(q) by Lemma 1. Assume that char
K 6= 2. We have a unique decomposition
z = �e1 � � � en + terms of lower length
by the P:B:W: theorem. In view of the expression for z computed between Lemma
4 and Proposition 5 of Chapter 5, the coeÆcient � is not zero (since char K 6= 2 !).
We get
C(')(z) = �'(e1) � � �'(en) + terms of lower length
= �det(')e1 � � � en + terms of lower length:
Since, on the other hand, C(')(z) = z + � for some ; � 2 K; 6= 0, we
must have = det(') by the uniqueness of the decomposition z = �e1 � � � en+terms of lower length. The formula C(')(z) = det(')z + � implies 2). We now
check 3) if (V; q) is 12{regular. Since Z(q) is graded and generated by an element
z of degree 1 and since C(') preserves grading, we must have C(')(z) = det(')z.
Thus C(') is the identity on Z(q) if and only if det(') = 1. If char K 6= 2, we
choose an orthogonal basis fe1; : : : ; eng for V . The element w = e1 � � � en is a gen-
erator of Z(q). For any ' 2 O(q), the basis fe0i = '(ei)g is also orthogonal and
C(')(w) = e01 � � � e0n = det(')w. This shows 3) if char K 6= 2.
Example 3. Let (V; q) = H(K). By Example 2 of Chapter 1,
O(q) = f(xu yv) 2M2(K) j xv + uy = 1; xu = 0; yv = 0g
if we identify V with K2 through a basis given by a hyperbolic pair fe; fg. The
element z = ef is a generator of Z(q). We have '(e) = xe + uf; '(f) = ye +
vf if ' = (xuyv): Thus
C(')(z) = (xe+ uf)(ye+ vf) = (xv � uy)z + uy:
If follows that
SO(q) =�
x 00 x�1
!2M2(K); x 2 K�
�' K�:
6. Special Orthogonal Groups and Spin Groups 61
In characteristic 2, O(q) \ SL2(K) = O(q), so SO(q) is a proper subgroup of
Ker(O(q)det! f�1g).
In the next section we assume that K is a �eld of characteristic not equal to
2. For any quadratic space (V; q) of dimension n, we de�ne the Lie algebra of the
group SO(q) as the set
so(q) = ff 2 EndK(V ) j bq(f(x); y) + bq(x; f(y)) = 0g;
where bq is the polar of q. If q = h1; : : : ; 1i, then so(q) = Altn(K). In particular
DimKso(h1; : : : ; 1i) = n(n�1)2
and the K{vector space so(q) is obviously a Lie algebra
for the Lie product [f; g] = fÆg�gÆf . Since charK 6= 2, any quadratic space (V; q) is
diagonalizable and there exists a �eld extension K � L such that L q ' h1; : : : ; 1i.Thus in general DimKso(q) =
n(n�1)2
and so(q) is a Lie algebra.
Let now [V; V ] be the K{linear subspace of C(q) generated by all [x; y] =
xy � yx; x; y 2 V . It follows from the P.B.W. theorem that DimK[V; V ] =n(n�1)
2.
Further, [V; V ] is a Lie algebra for the product [ ; ] and [[V; V ]; V ] � V . Again this
is easily seen by taking an orthogonal basis of V . Let
: [V; V ]! EndK(V )
be de�ned by (�) = [�; v] = �v � v�; � 2 [V; V ]. We have
bq([�; v]; w) + bq(v; [�; w]) = 0:
This can be checked for � = [x; y] using the de�ning relations of the Cli�ord algebra
and the general case follows by linearity. Thus is a homomorphism of Lie algebras
[V; V ] ! so(q). We claim that is an isomorphism. Since both algebras have the
same dimension, it suÆces to check that is injective. If (�) = 0, then � lies in the
centre of C(q) and in the centre of C0(q) so � 2 K. But K \ [V; V ] = 0. Therefore
the Lie algebra so(q) can be identi�ed with [V; V ] � C0(q). We observe that
[so(q); so(q)] = so(q) if n � 3:
This, again, can be easily checked by taking an orthogonal basis of V and us-
ing the identi�cation so(q) = [V; V ]. It follows that so(q) is always contained in
C0(q)0 = [C0(q); C0(q)]. We shall use this remark later.
62 6. Special Orthogonal Groups and Spin Groups
We now introduce di�erent subgroups of the group of units of the Cli�ord
algebra C = C0 � C1 of (V; q). These groups will help to describe O(q) and SO(q).
The �eld K can be of arbitrary characteristic. Let u 2 C� be a unit which is
homogeneous , i.e. u 2 C�0 or u 2 C�\C1. We de�ne the graded inner automorphism
igu : C ! C by igu(x) = (�1)@(u)@(x)uxu�1
for x homogeneous, @(u) denoting the degree of u, i.e. @(u) = 0 if u 2 C0 and
@(u) = 1 if u 2 C1. The group
�(q) = fu 2 C�; u homogeneous j igu(V ) � V g
is called the Cli�ord group of (V; q) and the subgroup
S�(q) = fu 2 C�0 j iu(V ) � V g = �(q) \ C0
is the special Cli�ord group.
Let � be the standard involution of C. We de�ne a map
� : C ! C by �(c) = �(c)c; c 2 C:
We have �(v) = �q(v) for v 2 V .
Lemma 4. Assume that (V; q) is nonsingular if the dimension of V is even and
12{regular if the dimension of V is odd. Then �(c) 2 K� for c 2 �(q) and � induces
a group homomorphism
� : �(q)! K�:
Proof. If suÆces to check that �(c) 2 K� for c 2 �(q). The property �(xy) =
�(x)�(y) then is immediate. For any c 2 �(q) and v 2 V , we have igc(v) = ��(igc(v)),since igc(v) 2 V . On the other hand �(igc(v)) = �ig�(c)�1(v), since �(v) = �v. There-fore �(c)cv = v�(c)c for all v 2 V . It follows that �(c)c is in the centre of C0 and in
the centre of C, so �(c)c 2 K�.
Example 5. An element v 2 V belongs to �(q) if and only if q(v) 2 K�, i.e., v is
anisotropic. For if q(v) 2 K�, then
igv(x) = �vxv�1 = �vxvq(v)�1 = (xv � bq(v; x))vq(v)�1
= x� bq(v;x)q(v)
v = �v(x):
6. Special Orthogonal Groups and Spin Groups 63
Thus igv is the re ection �v.
Since igu(V ) � V for u 2 �(q); igu 2 O(q) and we have a group homomorphism
�(q) ! O(q); u 7! igu. We denote it by �. Since iu is the identity on Z(q) if
u 2 S�(q); � restricts to a homomorphism S� : S�(q)! SO(q).
Proposition 6. The sequences
1! K� ! �(q)�! O(q)! 1
and
1! K� ! S�(q)S�! SO(q)! 1
are exact.
Proof. If igu(v) = v for all v 2 V , then u 2 ZC(C0) = Z(q) by Lemma 1 of Chapter
5. Thus u 2 S�(q) if DimV is even and igu = iu. But then iu(v) = v implies that
u 2 Z(C) = K. If the dimension of V is odd, either u 2 K� or u = �z; z a generator
of Z(q) as given in Proposition 5 of Chapter 5. If u = �z; we get igu(v) = �zvz�1,
thus zv = �vz. But zv + vz = v by Proposition 5 of Chapter 5, thus v = 0 which
is absurd. The surjectivity of � if (V; q) 6' H(IF2) ? H(IF2) follows from Exam-
ple 5 and the fact that O(q) is generated by re ections (Corollary 15 of Chapter
1). The case (V; q) = H(IF2) ? H(IF2) can be checked directly by counting the
order of the groups. We �nally verify that S� is always surjective. Let ' 2 SO(q).We have to show that C(') = iu for some u 2 C�
0 . If DimV is even and Z(q)
is a �eld, then C0 is c:s: over Z(q); C(')jC0 is Z(q){linear so by Skolem{Noether
C(')jC0 = iu; u 2 C�0 . If Z(q) = K �K, we apply Skolem{Noether componentwise
to get C(')jC0 = iu; u 2 C�0 . On the other hand C(') = iv for v 2 C� since C is
c:s:We have ivjC0 = iu, hence u = Æv for Æ 2 Z(q)� and we may assume that v 2 C0.
A similar argument works for the odd dimensional case since Z(q){automorphisms
of C are inner by the generalization of Skolem-Noether mentioned in Remark 12 of
Chapter 2.
Let ' 2 SO(q). By Proposition 6 there exists u 2 S�(q) such that C(') = iu
and by Lemma 4 �(u) 2 K�. We call the class of �(u) in K�=K�2 the spinor norm
64 6. Special Orthogonal Groups and Spin Groups
of ' and denote it by SN('). The element SN(') does not depend on the choice
of u such that C(') = iu. If iu = iv, then u = �v; � 2 K� and �(u) = �2�(v). Let
now
Spin(q) = fu 2 S�(q) j �(u) = 1g:
Theorem 7. The sequence
1! f�1g ! Spin(q)S�! SO(q)
SN! K�=K�2
is exact.
Proof. Obvious by the de�nition of S� and SN .
Even if its de�nition looks quite complicated, the group Spin(q) is very use-
ful. Moreover, Theorem 7 shows that Spin(q) is a good "approximation" of SO(q).
As we shall see later, for low dimensions, the group Spin(q) is easier to compute
than SO(q). As an algebraic group it is connected and simply connected if q is
of dimension � 3. There are also topological reasons to introduce Spin(q). If
(V; q) = (IRn; h1; : : : ; 1i); n � 3, then Spin(q) is the universal covering of SO(q).
Another classical group associated with a quadratic from (V; q) is its group of
similitudes. We denote it by GO(q). There is an exact sequence
1! O(q)! GO(q)�! K�;
where �('); ' 2 GO(q), is the multiplier of the similitude '. Let fe1; : : : ; eng be abasis of V and let � = (bq(ei; ej)) be the matrix of the polar of q. Further let ' be
the matrix of ' 2 GO(q). We have
't�' = ��
so �n = (det')2. Thus, if n is odd, � is a square, � = �2, and '0 = ��1' is an
isometry. It follows that there exists a direct product decomposition
GO(q) = O(q)�K� if n is odd:
By Lemma 10 of Chapter 5, similitudes of (V; q) induce automorphisms of C0. As-
sume that DimV is even. We say that a similitude ' is special if the induced
6. Special Orthogonal Groups and Spin Groups 65
automorphism C0(') of C0 is Z(q){linear, i.e. is the identity on the centre of C0.
We denote the group of special similitudes of (V; q) by GO+(q). It contains SO(q)
and K� but is not a direct product of these groups. The computation of GO+(q)
for nonsingular quadratic forms of dimension � 6 over a �eld of characteristic not
equal to 2 was done by Dieudonn�e in a paper of Acta Mathematica 1952 (see the
bibliography). In the next chapters, we shall study forms of dimension � 6 and in
particular reprove Dieudonn�e's results in a characteristic{free way.
Chapter 7
Quadratic Forms of Dimension 2
Let (V; q) be nonsingular of dimension 2 and let C = C0 � C1 be the Cli�ord
algebra of q. We have C1 = V; C0 = Z(q) and C is c:s: of dimension 4. Let � be
the standard involution of C (as a Cli�ord algebra) and �0 the standard involution
of Z(q). By Proposition 6 of Chapter 5, �0 is the restriction of � to C0. Further it is
easy to check, using a basis of C given by the Poincar�e{Birkho�{Witt theorem, that
�(x)x 2 K for all x 2 C. Hence � is the standard involution of C as a quaternion
algebra. Writing an element x 2 C as x = v + y; v 2 C1 = V and y 2 C0, we have
�(x) = �v + �0(y). Thus the reduced norm n of C is given by
n(x) = x�(x) = �q(v) + n0(y);
where n0 is the norm of C0 = Z(q), and we have an orthogonal decomposition
(C; n) ' (V;�q) ? (C0; n0):(7.1)
Proposition 1. Two quadratic spaces of dimension 2 are isometric if and only if
they have the same Witt invariants and the same Arf invariants.
Proof. Let q; q0 be the two forms and let C 0 = C 00 � V 0 be the Cli�ord algebra of
q0. From w(q) = w(q0), we get C ' C 0 and from a(q) = a(q0), we get C0 ' C 00. The
following lemma implies that (C; n) ' (C 0; n) and (C0; n0) ' (C 00; n0), so that the
claim is a consequence of Witt cancellation.
Lemma 2. Let A; B be algebras with standard involutions �A; �B. Any iso-
morphism ' : A ~!B of K{algebras is an isometry of the corresponding norms
nA(x) = x�A(x); nB(y) = y�B(y).
7. Quadratic Forms of Dimension 2 67
Proof. The involution '�A'�1 is a standard involution of B, hence by the unique-
ness of standard involutions '�A'�1 = �B. Thus
nB('(x)) = '(x)�B('(x)) = '(x)'(�A(x)) = '(nA(x)) = nA(x):
Example 3. Let Z be a separable quadratic algebra with standard involution �Z .
We compute the Cli�ord algebra of its norm. By de�nition, we have C(Z; nZ) =
(1; Z=K], where (1; Z=K] is the algebra Z�uZ with the multiplication rules u2 = 1
and xu = u�Z(x). We de�ne a K{linear map
� : Z ! EndK(Z)
by �(x)(y) = x�Z(y). Since �(x)2(y) = nZ(x)y, the map � induces a homomorphism
C(Z; nZ) ! EndK(Z), which is an isomorphism since C(Z; nZ) is c:s: over K. On
the other hand, if z 2 Z is a generator such that z2 = z + r and if we take f1; zg asa basis of Z, it is easy to check that C0(Z; nZ) ' Z. We get
C(Z; nZ) ' EndK(Z) and C0(Z; nZ) ' Z:
Thus, by Lemma 2 above and Lemma 10 of Chapter 5, two quadratic separable
algebras are isomorphic if and only if their norms are similar.
Proposition 4. Let (V; q) be a quadratic space of dimension 2. Then
1) (V; q) is isotropic if and only if Z(q) = C0 ' K �K, i.e. its Arf invariant is
trivial.
2) (V; q) represents 1 if and only if C(q) ' M2(K), i.e. its Witt invariant is
trivial. In particular, C(q) is a division algebra if and only if q does not represent 1.
Proof. 1) Assume that q represents some � 6= 0 2 K and let x 2 V with q(x) = �.
The map C0 ! C1 = V de�ned by c 7! xc is an isometry (C0; �n0) ~!(V; q) since
q(xc) = �n(xc) = �xc�(xc) = �x�(x)n0(c) = q(x)n0(c). Thus the claim follows
from Example 3 and the fact that q is hyperbolic if and only if �q is hyperbolic.
2) Assume that C ' M2(K). Then by Lemma 2,
(C; n) ' (M2(K); det) ' H(K2):
Further, by Proposition 4 of Chapter 1, (V; q) ? (V;�q) ' H(K2), so that (7.1)
and Witt cancellation imply that (V; q) ' (C0; n0). This shows that q represents
68 7. Quadratic Forms of Dimension 2
1. Conversely, if q represents 1, then (V; q) ' (C0; n0) and the claim follows from
Example 3.
Remark 5. The hyperbolic plane H(K) obviously represents any element � 2 K.
Hence if the Arf invariant of a quadratic space of dimension 2 is trivial, its Witt
invariant is trivial. Example 3 shows that the converse is not true.
Proposition 1 can be weakend to the following
Proposition 6. Two quadratic spaces of dimension 2 are similar if and only if they
have isomorphic even Cli�ord algebras, i.e. the same Arf invariants.
Proof. The claim follows from the fact that (V; q) ' (C0; �n0) if q represents the
element �.
We conclude this chapter by computing Spin(q) of a quadratic space (V; q) of
dimension 2. We use freely the notations of Chapter 6. The norm of C as a Cli�ord
algebra is the reduced norm of C, so that �(x) 2 K for all x 2 C. Since V = C1,
we have
S�(q) = C�0
and
Spin(q) = fx 2 C0 j n0(x) = 1g:Lemma 7. Let ' 2 SO(q). Then
1) C(')jC0 = 1C0 .
2) '(x) = ux; x 2 V , for some u 2 C�0 such that n0(u) = 1. Conversely, let
�u : x ! ux; x 2 V and u 2 C0, then �u 2 SO(q) if n0(u) = 1. In particular, the
map u 7! �u is an isomorphism
Spin(q) ~!SO(q):
Proof. 1) Follows from the de�nition of SO(q), since C0 is commutative. We prove
2). Let v 2 V be anisotropic and let q(v) = � 6= 0. We have for x 2 V; C(')(x) =
C(')(��1v2x) = ��1C(')(v)vx since vx 2 C0 and C0 is invariant under C('). The
element u = ��1C(')(v)v lies in C0 and '(x) = C(')(x) = ux. Since the map ' is an
7. Quadratic Forms of Dimension 2 69
isometry q(x) = q('(x)) = q(ux) = (ux)2 = n0(u)x2 = n0(u)q(x), hence n0(u) = 1.
Conversely let u 2 C�0 with n0(u) = 1, then �u 2 O(q). Since �u(x)�u(y) = u�(u)xy
for x; y 2 V , �u induces the identity on Z(q) and �u 2 SO(q).
The map S� : Spin(q)! SO(q) is given by u 7! iu. We have iu(x) = uxu�1 =
u�0(u)�1x by Proposition 6 of Chapter 5. Since n0(u) = 1; u�0(u
�1)
= u2 and S� can be identi�ed with u 7! �u2 :
Remark 8. By Example 3 any separable quadratic K{algebra is the even Cli�ord
algebra of a quadratic space of dimension 2. We claim that any c:s: K{algebra A
of dimension 4 is the Cli�ord algebra of a quadratic space of dimension 2. As in the
proof of Theorem 7, Chapter 3, we can �nd a separable quadratic K{algebra B � A
and an element u 2 B? � A (? with respect to the reduced norm of A) such that
A = B � uB; u2 = � 2 K� and ub = �B(b)u for b 2 B. By the universal property
of the Cli�ord algebra the map B ! A; b 7! (0; ub); extends to an isomorphism
C(B; �nB) ' A.
Chapter 8
Quadratic Forms of Dimension 3
Let (V; q) be a 12{regular form of dimension 3 with Cli�ord algebra C = C0�C1.
By Proposition 5 of Chapter 5, C0 is a c:s: K{algebra of dimension 4, hence a sep-
arable quaternion algebra, the centre Z(q) of C is a quadratic K{algebra generated
by an element z 2 C1 such that z2 = s 2 K and the class of s in K�=K�2 is the
signed 12{discriminant of q.
Example 1. Let A be a separable quaternion algebra (i.e. a c:s: K{algebra of
dimension 4) with standard involution �, trace tr, norm n and let
A0 = fx 2 A j x + �(x) = 0g:We claim that (A0; n), is 1
2{regular of dimension 3. By Remark 8 of Chapter 7, we can
assume that A is the Cli�ord algebra of a space [a; b]. If fx; yg is the correspondingbasis, then A0 has the basis f1 � 2xy; x; yg and (A0; n) ' h4ab � 1i ? [�a;�b] is12{regular, as claimed. The signed 1
2{discriminant of (A0; n) is the class of �1 in
K�=K�2. For A =M2(K), we get
A0 = fx 2M2(K) j tr(x) = 0g ' h�1i ? H(K):
We compute the Cli�ord algebra of (A0;�n). The signed 12{discriminant is trivial.
Let Z be the quadratic K{algebra generated by a symbol ~z such that ~z2 = 1. We
de�ne
' : A0 ! A Z
by '(a) = a ~z for a 2 A0. For any a 2 A, we have a2 � tr(a)a+ n(a) = 0, so that
a2 = �n(a) if a 2 A0 and '(a)2 = �n(a). By the universal property of the Cli�ord
algebra, ' induces a homomorphism (also called ')
' : C(A0;�n)! A Z:
8. Quadratic Forms of Dimension 3 71
Let fe0; e1; e2g be the basis f1; x; yg (if char K = 2), resp. fxy; x; yg (if char K 6= 2).
The element z = e0(1 � 2e1e2) of Proposition 5, Chapter 5, is mapped to 1 ~z so
that ' maps C0(A0;�n) into A. Since C0(A
0;�n) is c:s: over K; ' restricts to an
isomorphism C0(A0;�n) ' A and is an isomorphism C(A0;�n) ~!A Z. Thus
C(A0;�n) ' A Z; where Z = K[X]=(X2 � 1)
and
C0(A0;�n) ' A:
We now use Example 1 to describe (V; q) in general. Let C(q) = C0 Z(q) and
z 2 Z(q) as above. Since z is of degree 1, the homomorphism � : x 7! zx maps
V into C0. Let � be the standard involution of C as a Cli�ord algebra. Using a
basis of C0 given by the P:B:W: theorem, it is easy to check that x�(x) 2 K for any
x 2 C0. Hence � is the standard involution of C0 as a quaternion algebra. We get
�(zx) = �(x)�(z) = �xz = �zx for x 2 V , so that �(V ) � C 00. Since � is injective
and V; C 00 have the same dimension, � is an isomorphism V ~!C 0
0. Further, since
�n(zx) = (zx)2 = x2z2 = sq(x); � is an isometry
(V; q) ' (C 00;�sn):
We also have V = fx 2 C1 j �(x) + x = 0g since ��1 maps C 00 to the set of elements
x 2 C1 such that �(x) + x = 0. Summarizing we get
Proposition 2. Let (V; q) be a 12{regular form of dimension 3. Then
1) V = fx 2 C1 j x + �(x) = 0g, where � is the standard involution of C.
2) (V; q) ' (C 00;�sn), where C 0
0 = fx 2 C0 j x + �(x) = 0g and s 2 K is such
that [s] = 12disc(q) 2 K�=K�2.
Corollary 3. Two 12{regular forms of dimension 3 are isometric if and only if they
have same Witt invariants and same 12{discriminants. They are similar if and only
if they have same Witt invariants.
Proof. If suÆces to notice that if A; B are two quaternion algebras then, by unique-
ness of standard involutions, any homomorphism of algebras ' : A ! B maps A0
into B0.
72 8. Quadratic Forms of Dimension 3
Corollary 4. A 12{regular form (V; q) of dimension 3 has trivial Witt invariant if,
and only if, it is isotropic.
Proof. Since q is isotropic if and only if sq is isotropic for any s 2 K�, we can assume
that (V; q) = (A0; n) for some c:s: K{algebra A of dimension 4. Then w(q) = [A] in
Br(K). If [A] = 1, then A = M2(K), and A0 = s`2(K) is isotropic. Conversely, if
(A0; n) is isotropic, (A; n) is isotropic and A =M2(K).
Corollary 5. C0(q) is a division algebra if and only if (V; q) is anisotropic.
We now compute the special Cli�ord group, the spin group and the Lie algebra
of (V; q). Since the norm � of C0 is the reduced norm n of C0; �(x) 2 K for all
x 2 C0. Similarly, we claim that �(x) 2 K for all x 2 C1. Writing x = uz with
u 2 C0, we have �(x) = uz�(uz) = �z2�(u) 2 K. We now claim that iu(V ) � V
for all u 2 C�0 . By Proposition 2, V =
fx 2 C1 j x = ��(x)g: We get �(iu(x)) = �(uxu�1) = ��(u)�1x�(u) = �uxu�1 =
�iu(x) since u�(u) 2 K: This shows that
S�(q) = C�0 and Spin(q) = fu 2 C�
0 j n(u) = 1g:
For any c:s: K{algebra A, we denote the group of elements of Mn(A) of reduced
norm 1 by SLn(A),
SLn(A) = fx 2Mn(A) j nMn(A)(x) = 1g;
so Spin(q) = SL1(C0). By Proposition 2, we have (V; q) ' (C 00;�sn), where [s] =
12Æ(q) 2 K�=K�2, so that
SO(V; q) ' SO(V; sq) ' SO(C 00;�n):
We assume now that (V; q) = (A0;�n) for some quaternion algebra A. It follows
from Example 1 that
C0(V; q) = A and C = A Z
with Z generated by an element z of degree 1 such that z2 = 1. If we identify
V � C1 with zV � C0, then (V; q) is identi�ed with (A0;�n) as a subset of C0 = A.
8. Quadratic Forms of Dimension 3 73
The map
S� : SL1(A)! SO(A0;�n) = SO(A0; n)
is then given by u 7! iu, where iu is now viewed as an automorphism of C0 (not C).
Similarly
S� : S�(A0; n) = A� ! SO(A0; n)
is identi�ed with u 7! iu; u 2 A�. The map iu is indeed an isometry of A0 since
nA(uxu�1) = nA(x). Proposition 6 of Chapter 6 implies that
SO(q) ' C�0=K
�:
In particular
SO(A0; n) ' A�=K�:
If the 12{regular form (V; q) is anisotropic, C0 is a division quaternion algebra D and
Spin(q) ' SL1(D); S�(q) ' D�:
If (V; q) is isotropic, C0 'M2(K),
Spin(q) ' SL2(K); S�(q) ' GL2(K)
and
SO(q) ' PGL2(K):
Let for example (V; q) = (IR3; h1; 1; 1i) = (IH 0; n). We have S�(q) = IH�. Since
the standard involution of C0 = IH as a Cli�ord algebra is the standard involution of
IH as a quaternion algebra, we get SN(') > 0 for any ' 2 SO(IH 0; n) (by de�nition
SN(') = nIH(u), where C(') = iu). It then follows from Theorem 7 of Chapter 6,
that
S� : Spin(IH 0; n)! SO(IH 0; n)
is surjective. Therefore any ' 2 SO(IH 0; n) can be represented as '(x) = axa�1
with a 2 IH of reduced norm 1.
For the Lie algebra, we get
so(A0; n) = A0 = [A;A];
since so(q) � [C0; C0] and both algebras have the same dimension.
Chapter 9
Quadratic Forms of Dimension 4
Let (V; q) be a quadratic space of dimension 4 over K. The Cli�ord algebra
C = C0 � C1 is c:s: of dimension 16, the even Cli�ord algebra C0 is of dimension
8, its centre Z(q) is a separable quadratic K{algebra and C0 is c:s: of dimension 4
over Z(q) if Z(q) is a �eld or C0 ' A� A if Z(q) = K �K, A c:s: of dimension 4
over K.
Example 1. Let A be a quaternion algebra with standard involution a 7! a, norm
n and let � 2 K�. The map
'0 : A!M2(A); '0(a) =
0 a�a 0
!; a 2 A
induces a graded homomorphism ' : C(A; �n)!M2(A), which must be an isomor-
phism by Lemma 8 of Chapter 2. If we identify both algebras through ', we get
C0(A; �n) = A� A; Z(A; �n) = K �K and the standard involution � of C(A; �n)
is given by a bc d
!7!
a ���1c��b d
!:
since its restriction to '0(A) is �1. The canonical involution �0 corresponds to a bc d
!7!
a ��1c�b d
!:
Both involutions � and �0 are of even symplectic type, since they are the tensor
product of the standard involution of A with the involutions x yu v
!7!
x ���1u��y v
!=
1 00 ��
! x yu v
!t 1 00 ���1
!resp.
x yu v
!7!
x ��1u�y v
!;
9. Quadratic Forms of Dimension 4 75
which are of orthogonal type. Since there is a �eld extension K � L such that
L (V; q) ' H(L2) ' (M2(L); det);
the standard involution and the canonical involution of C(V; q); V a quadratic
space of dimension 4, are of even symplectic type. Since C(A; n) ' M2(A) and
Z(A; n) = K � K, the Witt invariant of (A; n) is the class of A in Br(K) and the
Arf invariant of (A; n) is trivial. Let A; B be quaternion algebras and
: (A; n)! (B; n)
a similitude with multiplier �. By the above computations and Lemma 10 of Chapter
5, induces an isomorphism
0 : A� A = C0(A; n) ~! C0(B; n) = B � B:
Since A and B are c:s: algebras, it follows that A ~! B. This, together with Lemma
2 of Chapter 7, shows that two quaternion algebras are isomorphic if and only if
they have similar norms.
The centre Z(q) of C0 is either a quadratic �eld extension ofK or Z(q) ' K�K.
In the �rst case C0 is a quaternion algebra over Z(q) and in the second C0 ' A�Aas a K�K{algebra (see Theorem 8 of Chapter 4). In both cases C0 has a standard
Z(q){linear involution (the componentwise involution in the second case).
Lemma 2. The standard Z(q){linear involution of C0 as a quaternion algebra is
the restriction to C0 of the standard involution � of C. Further, x�(x) 2 Z(q) for
any x 2 C1.
Proof. There exists a �eld extension K � L such that L (V; q) ' H(L2) '(M2(L); det). Thus by uniqueness of the standard involution, we may assume that
(V; q) = (A; n) for the algebra A = M2(K). The �rst claim then follows from the
formula for the involution given in example 1. For the second claim we may also
assume that (V; q) = (A; n). Let x = (0cb0) be an element of C1 = (0A
A0 ). We have
x�(x) =
��bb 00 ���1cc
!2
K 00 K
!= Z(q);
76 9. Quadratic Forms of Dimension 4
as claimed.
Let n : C1 ! Z(q) be the quadratic map de�ned by n(x) = x�(x). Even if Z(q)
is not necessarily a �eld, we can view Z(q) (V; q) in an obvious way as a quadratic
space over Z(q).
Lemma 3. The multiplication in C induces an isometry : Z(q) (V; q) '(C1;�n). In particular Z(q) (V; q) is similar to (C0; n).
Proof. Let y 2 Z(q) and v 2 V . Since n(yv) = yv�(yv) = �y2q(v), the map
is a morphism of quadratic spaces. Further is injective because Z(q) (V; q)
is nonsingular. Comparing dimensions shows that is an isomorphism. The last
claim follows by choosing some v 2 V anisotropic and observing that the map
c 7! vc; c 2 C0, is a similitude (C0; n)! (C1; n) with multiplier �q(v) 6= 0.
Assume that the Arf invariant of (V; q) is not trivial, i.e. Z(q) is a quadratic
�eld extension L of K. It follows from Lemma 3 that C0 is a quaternion division
algebra over L if and only if L(V; q) is anisotropic and that C0 ' M2(L) if and only
if L (V; q) is isotropic. On the other hand we claim that L (V; q) is anisotropic if
and only if (V; q) is anisotropic. Assume that L (V; q) is isotropic and that (V; q)
is anisotropic. Let z be a generator of L such that z2 = z + r; r 2 K and let
u+ zv 6= 0 2 L V; u; v 2 V be such that q(u+ zv) = 0. We have
q(u) + rq(v) = 0 and bq(u; v) + q(v) = 0:
Let s = q(v); s 6= 0 since q is anisotropic. We get bq(u; u) bq(u;�v)bq(�v; u) bq(�v;�v)
!=
�2rs ss 2s
!
and det(�2rss
s2s) = �s2(1 + 4r) 6= 0. Therefore u; �v are linearly independent and
qjU ; U = Ku�K(�v), is nonsingular. The form qjU is isometric to (L; sn) where
n is the norm of L, so there exists an orthogonal decomposition
(V; q) ' (L; sn) ? (V 0; q0):
By Example 3 of Chapter 7 and Lemma 10 of Chapter 5, the Arf invariant of (L; sn)
is L. Since L is also the Arf invariant of (V; q); (V 0; q0) must have trivial Arf invari-
9. Quadratic Forms of Dimension 4 77
ant (consider the cases char K 6= 2 and char K = 2 separately). By Proposition 4
of Chapter 7, (V 0; q0) is isotropic, hence (V; q) is isotropic. This is a contradiction.
Summarizing, we get
Proposition 4. Let (V; q) be a quadratic space of dimension 4 with nontrivial Arf
invariant L. Then
1) C0(q) is a quaternion division algebra over L if and only if (V; q) is anisotropic.
2) C0(q) 'M2(L) if and only if (V; q) is isotropic.
Proposition 5. V = fx 2 C1 j x+ �(x) = 0g.
Proof. Let �0 be the standard involution of Z(q) and let ~� = (�0 1) �1, with
as in Lemma 3. We get
V = fx 2 C1 j ~�(x) = xg
by Lemma 3 (and Galois descent!). But ~�(yv) = �0(y)v = vy = ��(yv) by Propo-
sition 6 of Chapter 5, thus ~� = �� and V = fx 2 C1 j x = ��(x)g as claimed.
We now put Z(q) = Z and identify Z V with C1. Let a; b 2 C0 and x 2 C1.
The map x 7! ax�(b) is a homomorphism
: C0 Z �C0 ! EndZ(Z V ) = Z EndK(V ):
Lemma 6. 1) is an isomorphism and restricts by Galois descent to an isomorphism
: cor(C0) ~!EndK(V ):
2) The involution cor(�) of cor(C0), induced by the standard involution � of
C0, is such that
cor(�) �1(f) = h�1q f �hq; f 2 EndK(V );
where hq : V ~!V � is the adjoint of q and f � is the transpose of f .
Proof. The map is an isomorphism, since it is a map between c:s: algebras of the
same dimension over Z. We prove that cor(C0) ' EndK(V ). Let f = (Pai bi) 2
EndK(V ). It suÆces to check that f = (Pbi ai). Putting x = �(x), we have, by
78 9. Quadratic Forms of Dimension 4
Proposition 5, f(v) = �f(v) for v 2 V . Since f(v) = Paibi = �P biai = �f(v),
we get f = (Pbi ai) as claimed. To prove 2), we verify that
( (a b))� Æ (1 hq) = (1 hq) Æ (a b)(9.1)
for a; b 2 C0. We have
(1 hq)(x)(y) = bq(x; y) = �(x y + y x)
by Lemma 3, so that (9.1) reduces to
x�(ay b) + ay b �(x) = a xb�(y) + y�(a xb)(9.2)
for a; b 2 C0; x; y 2 C1. The left hand side of (9.2) is equal to tr(xb y a) and the
right hand side to tr(axby), where tr is the reduced trace of C0 as Z{algebra. Both
sides are equal since tr(uv) = tr(vu) for u; v 2 C0.
Theorem 7. Two quadratic spaces of dimension 4 over K are similar if and only if
they have isomorphic even Cli�ord algebras.
Proof. If q; q0 are similar, then C0(q) ' C0(q0) by Lemma 10 of Chapter 5.
Conversely, let ' : C0(q) ~! C0(q0) be an isomorphism of K{algebras and let Z =
Z(q) be the centre of C0(q). Then ' induces an isomorphism Z ~!Z(q0). We view
C 00 = C0(q
0) as a Z{algebra through this isomorphism, so that ' is Z{linear. In view
of the uniqueness of standard involutions, ' is an isomorphism of Z{algebras{with{
involution. By Lemma 6, ' induces an isomorphism : EndK(V )! EndK(V0) such
that Æ = 0 Æ cor('), where : cor(C0) ~!EndK(V ); 0 : cor(C 0
0) ~!EndK(V0)
are as in Lemma 6. Hence is an isomorphism of algebras{with{involution. Fixing
isomorphisms EndK(V ) ' M4(K); EndK(V0) ' M4(K), we can write by Skolem{
Noether, (f) = �f��1 for some K{linear isomorphism � : V ~!V 0. Since is an
isomorphism of algebras{with{involution, we get
�h�1q f �hq�
�1 = h�1q0 (� f �
�1)�hq0
for all f 2 EndK(V ). This implies that � � hq0 = ��hq� for some � 2 K�. Therefore
� is a similitude (V; bq) ~!(V 0; bq0) of the polar forms. To check that � is in fact a
similitude of the quadratic spaces, we observe that
9. Quadratic Forms of Dimension 4 79
1) the reduced trace tr : C0 ! Z induces an isomorphism
t : HomC0(Z M;C0)! HomZ(Z M;Z);
where we view Z M = C1 as a left C0{module through multiplication in C.
2) t�1 Æ (1 hq)(v)(v) = 1 q(v) for v 2 V .3) The reduced trace commutes with '.
3) follows by uniqueness of the standard involution. 1) and 2) can easily be checked
in the case (V; q) = (A; n) (see Example 1) and the general case follows from the
fact that there exists a �eld extension K � L such that L (V; q) ' H(L2) =
(M2(L); det).
Corollary 8. Two quadratic spaces of dimension 4 are similar if they have the same
Witt and Arf invariants.
Proof. By Lemma 9 of Chapter 4, we have [C0(q)] = [Z(q) C(q)] in Br(Z(q))
(assume that Z(q) is a �eld or do the necessary changes if Z(q) = K �K !). There-
fore C0(q) ' C0(q0) if Z(q) ' Z(q0) and [C(q)] = [C(q0)] in Br(K).
Not any multiplier � can occur in Corollary 8. We have:
Lemma 9. Let (V; q) be a quadratic space of even dimension. Then w(q) = w(�q)
if and only if � = n0(�); � 2 Z(q)� and n0 is the norm of Z(q).
Proof. If � = n0(�), then V ! C(q) given by v 7! �v extends to an isomorphism
C(�q) ~!C(q). If w(q) = w(�q), then w((�; Z(q)=K]) = 1 by Proposition 11 of
Chapter 5. The map
Z(q)! (�; Z(q)=K] = Z(q)� uZ(q)
given by x 7! (0; ux) extends by the universal property of the Cli�ord algebra to an
isomorphism
C(Z(q); �n0) ' (�; Z(q)=K]:
By Proposition 4 of Chapter 7, C(Z(q); �n0) ' M2(K) if and only if �n0 represents
80 9. Quadratic Forms of Dimension 4
1, i.e. n0 represents �.
Remark 10. Let Z be a quadratic algebra and let B be a quaternion algebra over
Z. By Lemma 6 a necessary condition for B to be isomorphic to the even Cli�ord
algebra of a quadratic space of dimension 4 is that the class of the corestriction of
B in Br(K) is trivial. As shown in Knus{Parimala{Sridharan this condition is also
suÆcient.
Remark 11. As an application of Theorem 7, (and Example 1) we get that a
quadratic space of dimension 4 and trivial Arf invariant is similar to the norm of a
quaternion algebra. This could of course be easily checked directly.
To discuss the structure of C (in particular to give conditions under which C is
a division algebra) we need some results of Chapter 11. We shall see in Proposition
8 of Chapter 11 that C is a division algebra if and only if (Z(q); n0) ? (V;�q) isanisotropic.
We now compute the groups Spin(q), the Lie algebra so(q) and GO+(q) for
quadratic spaces of dimension 4. By de�nition the group Spin(q) is a subgroup of
the group SL1(C0) of units of C0 of Cli�ord norm 1. We claim that in fact
Spin(q) = SL1(C0):
We have to check that for any u 2 SL1(C0); iu(V ) � V . By Proposition 4, V =
fx 2 C1 j x+ �(x) = 0g. We get
�(iu(x)) = �(uxu�1) = ��(u�1)x�(u) = �uxu�1;
since u�(u) = 1. Thus �(iu(x)) = �iu(x) as claimed.
Assume now that (V; q) has trivial Arf invariant. As observed in Remark 11,
(V; q) is similar to the reduced norm of a quaternion algebra A. Therefore, to
compute Spin(q) and SO(q), we may as well assume that (V; q) = (A; n). As usual,
we denote by a 7! �a the standard involution of A. By Example 1,
C0 =
A 00 A
!�M2(A) = C
9. Quadratic Forms of Dimension 4 81
or C0 = A� A. The involution � maps (a; b) to (a; b). Since
V =�
0 xx 0
!; x 2 A
�;
the condition iu(V ) � V for u 2 C�0 is equivalent to bxa�1 = ax b�1 for all x 2 A.
This, in turn, is equivalent with n(a) = n(b). Thus
S�(q) = f(a; b) 2 A� A j n(a) = n(b)g;
Spin(q) = f(a; b) 2 A� A j n(a) = n(b) = 1g = SL1(A)� SL1(A)
and the homomorphism S� : S�(q) ! SO(q), resp. Spin(q) ! SO(q) is given by
S�(a; b)(x) = bxa�1. If A =M2(K), we get
Spin(H(K2)) = SL2(K)� SL2(K):
If A = IH, the spinor norm of any ' 2 SO(IH; n) is trivial, so that by Theorem
7 of Chapter 6, any ' 2 SO(IH; n) can be represented as '(x) = bxa�1 for a; b
quaternions of norm 1. In general, by Proposition 6 of Chapter 6 and the above
computation of S�(q), any ' 2 SO(A; n) can be written as '(x) = bxa�1 for ele-
ments a; b 2 A of equal norm. We recall that A = M2(K) if (V; q) is isotropic and
A is a division algebra if (V; q) is anisotropic.
If the Arf invariant of (V; q) is not trivial, the centre of C0(q) is a quadratic
�eld extension L of K and C0 is a quaternion algebra over L. Thus C0 is either a
division algebra over L or C0 ' M2(L). By Proposition 4 the �rst case occurs if
(V; q) is anisotropic and the second if (V; q) is isotropic. Therefore
Spin(V; q) ' SL2(L)
if (V; q) is isotropic and
Spin(V; q) ' SL1(B);
with B = C0 a division algebra over L if (V; q) is anisotropic. We claim that
B ' L B0;
B0 a division algebra over K. By Lemma 5, cor(B) is trivial. By the theorem of
Albert{Riehm (Theorem 11 of Chapter 3), B admits an involution � of the second
kind, i.e. such that its restriction to L is the standard involution �0 of L. Now ���
82 9. Quadratic Forms of Dimension 4
is an involution of B which is L{linear and such that ���(x) � x 2 L for all x 2 B;since ��(x) � �(x) 2 L. In view of the uniqueness of the standard involution �,
we get ��� = � or (��)2 = 1. The map ~� = �� : B ! B is �0{semilinear.
By Galois descent, we have B ' L B0 with B0 = fx 2 B j ~�(x) = xg. One
could also construct B0 such that B = L B0 using a basis of V : Assume that
(V; q) = [a; b] ? [c; d] and let fx1; y1; x2; y2g be a corresponding basis of V . Then
z = x1y1 + x2y2 � 2x1y1x2y2
is a generator of L and the elements x1y1; x1x2; y1x2 generate a subalgebra B0 of
dimension 4 over K. It can be checked that L B0 = C0, so B0 is c:s: over K.
The computation of the Lie algebra so(q) is simple. We have so(q) � C 00 =
[C0; C0] in general. Since
C 00 = fx 2 C0 j x+ tr(x) = 0g
is of dimension 3 over Z, we must have
so(q) = C 00 = fx 2 C0 j x + tr(x) = 0g:
We �nally compute the group of special similitudesGO+(q) of a quadratic space
of dimension 4. Let
� : K� � C�0 ! GO(q)
be de�ned by �(�; u)(x) = ��1uxu. Since M = fx 2 C1 j x + �x = 0g (see (4.2.3)),
�(�; a)(x) is a linear automorphism of V for all a 2 C�0 ; � 2 R�. Further it follows
from
q(�(�; a)(x)) = (�(�; a)(x))2
= ��2ax�aax�a= ��2a�0(�(a))q(x)�a= ��2n0(�(a))q(x)
that �(�; a) is a similitude with multiplier ��2n0(�(a)), where �(a) = a�a 2 Z (see
(4.1.1)). We now check that �(�; a) 2 GO+(q). We have for all elements x; y of V
�(�; a)(x)�(�; a)(y) = ��2ax�aay�a= ��2a�0(�(a)xy�a= ��2�(a)�0(�(u)axya
�1
= ��2n0(�(a))axya�1
9. Quadratic Forms of Dimension 4 83
Since the products xy; x; y 2 V , generate the algebra C0, we see that the extension
C0(�(�; a)) to C0 of the similitude �(�; a) is the inner conjugation ia by a. Thus
C0(�(�; a)) is Z{linear and, by de�nition, �(�; a) is a special similitude. We now
claim that the kernel of � is the subgroup
f(n0(z); z) 2 K� � Z�j z 2 Z�g
of K� � C�0 . It follows from �(�; a)) = 1 that C0(�(�; a)) = ia = 1C0 and a is an
element of the centre Z of C0. We have a = �a for a 2 Z. Thus �(�; a)(x) = x
implies ��1a�0(a)x = ��1n0(a)x = x for all x 2 V . Since V is nonsingular, we get
n0(a) = � as claimed. We �nally claim that � is surjective. Let Z = Z(q). For any
Z{module M , let �0M be the Z{module M with the action of Z twisted through
�0. The subset E of M2(C) consisting of matrices a bc d
!2
�0C0 �0C1
C1 C0
!
is a Z{algebra. The map
Z V ! E; x 7!
0 xx 0
!;
extends by the universal property of the Cli�ord algebra to a homomorphism Z C ! E, which is an isomorphism since both algebras have the same dimension and
C is c:s: over K. We use it to identify Z C with E. The standard involution of
Z C is
� :
a bc d
!7! a cb d
!
since � restricted to Z V = f(0x x0); x 2 Z V g is �1ZV . Let f 2 GO+(q) with
multiplier �. The map 0 �xx 0
!7!
0 ��1f(x)
f(x) 0
!; x 2 V;
extends by the universal property of the Cli�ord algebra to an automorphism ' of
Z C. It is easy to check that the restriction '0 of ' to
Z C0 =
�C0 00 C0
!
is the automorphism C0(1 f). By Skolem{Noether (and Remark 12 of Chap-
ter 2), ' = iu for some u 2 (Z C)�. Since f 2 GO+(q); '0 = C0(1 f) is
84 9. Quadratic Forms of Dimension 4
Z Z{linear, hence we also have '0 = iv; v 2 (Z C0)�. Since iv = iu on
ZC0 u = rv; r 2 ZZ = (Z00Z) and u 2 (ZC0)
�. Let u = (a00d) 2 (ZC0)
�,
so that f(x) = dxa�1. Since V = fx 2 C1 j x + x = 0g we get dxa�1 = a�1xd for
all x 2 V or cx = xc with c = ad. It follows that cx 2 V for all x 2 V . Thus we get(cx)2 = cxx�c = c�cq(x) 2 K� for all x 2 V , so that c�c 2 K�. Since also �cxc 2 V , wehave c�cxc = �cxc�c, hence xc = �cx for all x 2 V . It follows from xyc = x�cy = cxy that
c 2 Z and, since �c = c for c 2 Z, that xc = cx for all x 2 V . This �nally implies
that c 2 K�, let c = �, so that a = �d�1
and f(x) = ��1dxd for d 2 C0, as claimed.
Summarizing, we get
Theorem 12. Let (V; q) be a quadratic space of dimension 4 with Cli�ord algebra
C = C0�C1. The homomorphismK��C�0 ! GO(q); (�; u) 7! �(�; u)(x) = ��1uxu
induces an isomorphism
K� � C�0=f(n0(�); �); � 2 Z� g ~!GO+(q);
where Z is the centre of C0.
We describe some special cases of Theorem 12. If (V; q) has trivial Arf invariant,
we have GO+(q) ' GO+(A; n) for some quaternion algebra A, since (V; q) is similar
to (A; n). Then
C�0 = A� � Z�; Z� = K� �K� and n0(Z
�) = K�;
so
GO+(q) ' A� � A�=f(�; �); � 2 K�g:
In particular
GO+(q) ' GL2(K)�GL2(K)=f(�; �); � 2 K�g
if further (V; q) is isotropic, i.e. (V; q) = H(K2).
If the Arf invariant of (V; q) is not trivial, Z is a quadratic �eld extension of K and
Z (V; q) is similar to a quaternion algebra (B; n) over Z. Then
GO+(q) = K� � B�=f(n0(z); z); z 2 Z�g;
9. Quadratic Forms of Dimension 4 85
where B is a division algebra if (V; q) is anisotropic and B = M2(Z) if (V; q) is
isotropic. In this case
GO+(q) = K� �GL2(Z)=f(n0(z); z); z 2 Z�g:
Chapter 10
The PfaÆan
We �rst recall the de�nition and some properties of the pfaÆan of an alternating
matrix and then de�ne a reduced pfaÆan for c:s: algebras. Let xij; 1 � i < j � 2m
be indeterminates and let xji = �xij for i < j; xii = 0; i = 1; : : : ; 2m. The matrix
� = (xij) with entries in ZZ[xij] is called the generic alternating (2m� 2m){matrix.
In view of Lemma 2 of Chapter 3 its determinant is a square in IQ(xij). Since det� 2ZZ[xij] and ZZ[xij] is integrally closed in IQ(xij), det� is the square of a polynomial
pf(�) in ZZ[xij]. The polynomial pf(�) is uniquely determined up to a factor �1.We normalize pf(�) by requiring that pf(sm) = 1 for sm = diag(( 0
�110); : : : ; (
0�1
10)).
In particular, pf(�) is homogeneous of degree m, i.e. pf(��) = �mpf(�) for � 2 K.
For example, we get
pf
0BBB@0 x12 x13 x14
0 x23 x240 x34
0
1CCCA = x12x34 � x13x24 + x14x23:
Let now K be a �eld and ZZ[xij]! K[xij ] be induced by the unique homomorphism
ZZ ! K. Specializing � 7! � for any alternating (2m� 2m){matrix �, we get
det(�) = pf(�)2:
Lemma 1. For any � 2 Alt2m(K) and � 2M2m(K), we have
pf(�t��) = det(�)pf(�):
Proof. It suÆces to verify the claim over ZZ for � and � generic matrices (i.e. with
indeterminates aij for � and uij for � as entries). We have pf(�t��)2 = det(�t��) =
det(�)2pf(�)2 over the polynomial ring ZZ[aij; uij], so that pf(�t��) = �det(�)pf(�)
and the sign � is independent of the choice of � and �. Specializing � to the identity
10. The PfaÆan 87
matrix shows that the sign + must hold.
Let A be a c:s: K{algebra of dimension 4m2 with a K{linear involution � . We
assume that either � is of even symplectic type or that A is isomorphic as an algebra
with involution to the tensor product of two quaternion algebras with the involution
given by the tensor product of the two standard involutions. For example M4(K),
with the involution x 7! xt, is isomorphic to such a tensor product. Let " be the
type of � and let
Alt� (A) = fx� "�(x); x 2 Ag
be the set of alternating elements of A. We claim that there exists a reduced pfaÆan
pfA : Alt� (A)! K such that
pfA(x)2 = nA(x) and pfA(�(a)xa) = nA(a)pfA(x)
for x 2 Alt� (A); a 2 A, nA denoting the reduced norm of A. We �rst consider
the case where � is of even symplectic type. Let � : L A ~!M2m(L) be a Galois
splitting of A. The involution �(1 �)��1 of M2m(L) is of the form �u(x) = uxtu�1
with u 2 GL2m(K) an alternating matrix. In view of Lemma 2 of Chapter 3, we can
modify � by an inner automorphism of M2m(L) in such a way that u is the matrix
sm de�ned above.We get by Lemma 1 of Chapter 3,
�(L Alt� (A)) = Alt�u2m(L)
= sm � Alt2m(L) = Alt2m(L) � s�1m
and we de�ne
pfA(a) = pf(�(1 a)sm) for a 2 Alt� (A):
We �rst check that pfA(a) 2 K of all a 2 Alt� (A). For any g 2 Gal(L=K), let
~g : M2m(L) ! M2m(L) be given by ~g((aij)) = (g(aij)) i.e. ~g acts entrywise. The
map �(g1)��1~g�1 is an L{automorphism ofM2m(L) which respects the involution
�(1 �)��1 of M2m(L). Hence there is c 2 GL2m(L) such that
�(g 1)��1(x) = c~g(x)c�1 for all x 2M2m(L)
and �sm = csmct for some � 2 L�. We have
pfA(a) = pf(�(1 a)sm) = pf(c~g(�(1 a))c�1sm)
88 10. The PfaÆan
since �(g 1)��1(�(1 a)) = �(1 �). On the other hand c�1sm = ��1smct, so
that
pfA(a) = pf(��1c~g(�(1 a))smct)
= ��mdet(c)pf(~g(�(1 a))sm):
We get �m = det(c) by taking the pfaÆan of �sm = csmct and
pf(~g(�(1 a))sm) = pf(~g(�(1 a)sm))
= g(pf(�(1 a)sm)) = g(pfA(a));
since sm is de�ned over K. This shows that pfA(a) 2 K. To verify that pfA is
independent of the choice of the splitting (L; �), it suÆces (as for the characteristic
polynomial) to consider splittings (L; �) and (L; �0) (i.e. the same �eld L). Let
�0 = iy Æ �; y 2 GL2m(L). Since �(1 �)��1 and �0(1 �)�0�1 give the involution
�sm ofM2m(L), we get by the discussion between Lemma 1 and Lemma 2 of Chapter
3, �sm = ysmyt for some � 2 L�. Let now pfA be the pfaÆan de�ned through �
and pf 0A the pfaÆan de�ned through �0. We have
pf 0A(a) = pf(�0(1 a)sm) = pf(y�(1 a)y�1sm)
= pf(��1y�(1 a)smyt) = ��mdet(y)pf(�(1 a)sm)
= pfA(a)
since �sm = ysmyt implies �m = det(y) (take the pfaÆan on both sides). The
formulas
pfA(x)2 = nA(x) and pfA(�(a)xa) = nA(a)pfA(x)
follow from
pfA(x)2 = pf(�(1 x)sm)
2 = det(�(1 x)sm) = det(�(1 x)) = nA(x)
and
pfA(�(a)xa) = pf(sm�(1 a)ts�1m �(1 x)�(1 a)sm)
= pf((s�1m �(1 a)sm)
t�(1 x)sm(s�1m �(1 a)sm))
= det(�(1 a))pf(�(1 x)sm)
= nA(a)pfA(x):
10. The PfaÆan 89
Further we have pfA(�a) = �mpfA(a) for � 2 K and pfA(1) = pf(sm) = 1 (we
observed earlier that 1 2 Alt� (A) if � is of even symplectic type).
We now de�ne the reduced pfaÆan pfA for A = A1 A2; A1; A2 quaternion
algebras and � = �1 �2 the tensor product of the standard involutions �i of Ai.
We observe that � is of orthogonal type since �1; �2 are of symplectic type. Let
�i : L Ai ~!M2(L) be a splitting of Ai; i = 1; 2 (we can choose the same L for
both algebras). Then � = �1 �2 is a splitting of A and �(1 �)��1 is the tensor
product of two copies of the involution
� = (xuyv) 7!
�v�u
�yx
�= s1�
ts�11 :
Thus �(L Alt� (A)) = (s1 s1)Alt4(L) = Alt4(L)(s1 s1)�1, where
s1 s1 =
0BBB@0 0 0 10 0 �1 00 �1 0 01 0 0 0
1CCCAand we de�ne
pfA(a) = pf(�(1 a)(s1 s1)) for a 2 Alt� (A):
As in the case of even symplectic involutions, one checks that pfA(a) 2 K for all
a 2 Alt� (A) and that pfA does not depend on the choice of the splitting �
(� always of the form �1 �2 !) One has to use that det(u1 u2) = (detu1 � detu2)2for ui 2 GL2(L). We leave the details as an exercise.
Let now A be a c:s: algebra of dimension 16 with an involution � which is either
of even symplectic type or A = A1 A2 and � = �1 �2; Ai a quaternion algebra
with standard involution �i; i = 1; 2.
Lemma 2. pfA : Alt� (A)! K is a nonsingular quadratic form on the vector space
Alt� (A) of dimension 6.
Proof. By de�nition of the reduced pfaÆan, it suÆces to consider the case of the
classical pfaÆan on Alt4(K). The formula given at the beginning of this chapter
shows that
(Alt4(K); pf) ' H(K3):
90 10. The PfaÆan
We now construct a \standard involution" on Alt� (A); A as above. We �rst
consider the case A =M4(K). Let � : Alt4(K)! Alt4(K) be de�ned by
�
0BBB@0 x12 x13 x14
�x12 0 x23 x24�x13 �x23 0 x34�x14 �x24 �x34 0
1CCCA =
0BBB@0 �x34 x24 �x23x34 0 �x14 x13�x24 x14 0 �x12x23 �x13 x12 0
1CCCA :We get x�(x) = �(x)x = pf(x) for x 2 Alt4(K). Further � is, in some sense, unique:
Lemma 3. Let : Alt4(K) ! Alt4(K) be a K{linear automorphism such that
(x)x 2 K for all x 2 Alt4(K) (or x (x) 2 K for all x 2 Alt4(K)). There exists
� 2 K� such that = ��.
Proof. Let fEijg; 1 � i; j � 4 be the standard basis of M4(K). Let
fgig; 1 � i � 6 be the basis fhij = Eij �Ejig of Alt4(K) ordered lexicographically.
The condition (g1)g1 2 K implies that (g1) = �g6; � 2 K and (g6)g6 2 K
implies (g6) = �0g1; �0 2 K. The condition (g1 + g6)(g1 + g6) 2 K then implies
� = �0 and the lemma follows by further similar computations.
If A is c:s: of dimension 16 as above, we shall de�ne �A : Alt� (A) ! Alt� (A)
through a splitting � : L A ~!M4(L) as above. We put
�(1 �A(a)) = s2�(�(1 a)s2) a 2 A
in the even symplectic case and
�(1 �A(a)) = (s1 s1)�(�(1 a)(s1 s1))
if A = A1 A2. Let s = s2 (resp. s1 s1). We have
�(1 a)�(1 �A(a)) = �(1 a)s�(�(1 a)s)
= pf(�(1 a)s) = pfA(a);
�(1 �A(a))�(1 a) = s�(�(1 a)s)�(1 a)
= s(�(�(1 a)s)�(1 a)s)s�1
= spf(�(1 a)s)s�1 = pfA(a)
so
�A(a)a = a�A(a) = pfA(a); a 2 Alt� (A):
10. The PfaÆan 91
Further �A is, up to scalars, the unique isomorphism : Alt� (A) ! Alt� (A) such
that (a)a 2 K for all a 2 Alt� (A) (or such that a (a) 2 K for all a 2 Alt� (A)).
The automorphism �A allows us to compute the Cli�ord algebra of (Alt� (A); pfA):
Proposition 4. Let � 2 K�. The map
Alt� (A)!M2(A); x 7!
0 �A(x)�x 0
!induces an isomorphism ' : C(Alt� (A); �pfA) ~!M2(A).
Proof. The existence of ' follows from the universal property of the Cli�ord alge-
bra. It is an isomorphism since C(Alt� (A); pfA) is c:s: and both algebras have the
same dimension.
If we identify C(Alt� (A); pfA) with M2(A), we get C0 = (A00A); C1 = (0A
A0 )
and the standard involution � of C is given by
�
a bc d
!=
d "b"c a
!=
0 "1 0
! d bc a
! 0 "1 0
!where a 7! a = �(a) is the involution of A and " is the type of � . Indeed, the
restriction of � to
V '�
0 �A(x)�x 0
!; x 2 Alt� (A)
�is �1V .
Let A be c:s: over K with an even symplectic involution � and let A[X] =
A K[X] be the polynomial ring in one variable X over A. We extend � to A[X]
by putting �(X) = X. We obviously have
Alt� (A[X]) = Alt� (A)K K[X]
and we get by scalar extension a pfaÆan pfA[X] over A[X] with values inK[X]. Since
1 2 Alt� (A), we can de�ne, for all a 2 Alt� (A), a polynomial ��(X; a) 2 K[X] by
��(X; a) = pfA[X](X � 1� a):
We call ��(X; a) the pfaÆan characteristic polynomial. It is of degree m if A has
dimension 4m2. In particular, if m = 2,
��(X; a) = X2 � pft(a)X + pfA(a)
92 10. The PfaÆan
where pft is a linear form on Alt� (A). We call it the pfaÆan trace of A. We have a
Cayley{Hamilton theorem. i.e. ��(a; a) = 0 for all a 2 Alt� (A).
Example 5. Let B be a quaternion algebra with standard involution b 7! b and let
A =M2(B). The involution
� :
a bc d
!7!
a cb d
!
of A is of even symplectic type since it is the tensor product of the involution x 7! xt
of M2(K) with the standard involution of B. We get
Alt� (A) = f�xb
by
�; x; y 2 K; b 2 Bg:
To compute the pfaÆan, we split B and apply the formula pfA(a) =
pf(�(1 a)sm) (which de�nes pfA(a) !). We get
pfA�xb
by
�= x y � bb:
Let : Alt� (A)! Alt� (A) be de�ned by �xb
by
�=�y�b
�bx
�: Since a (a) = pfA(a),
we must have = �A, so
�A
x bb y
!=
y �b�b x
!:
Further pft(a) = b(a; 1), where b is the polar of pfA, so
pft
x bb y
!= x+ y
(as one would guess!). In particular, we get
a+ �A(a) = pft(a) 2 K for all a 2 Alt� (A):
This strengthens the (heuristic) fact, already mentioned, that �A is a "standard
involution" on Alt� (A).
Chapter 11
Quadratic Forms of Dimension 6
At the end of Chapter 10, we have shown that the Arf invariant of a quadratic
space (Alt� (A); �pfA); A a c:s: K{algebra of dimension 16 with an even symplectic
involution and � 2 K�, is trivial. Conversely we claim that
Proposition 1. Let (V; q) be a quadratic space of dimension 6 with trivial Arf
invariant. Assume that (V; q) represents � 2 K�. There exist a c:s: K{algebra A
and an even symplectic involution � on A such that (V; q) ' (Alt� (A); �pfA).
Let C = C0�C1 be the Cli�ord algebra of (V; q) and let e; f be two nontrivial
idempotents generating Z(q) = K �K. Before proving Proposition 1, we describe
V as a subspace of C1. We denote as usual the standard involution of C by �.
Lemma 2. The map V ! V e; x 7! xe, is an isomorphism and
V e = f(x� �(x))e; x 2 C1g = fy � �(y); y 2 C1eg:
Similarly V ' V f and
V f = f(x� �(x))f; x 2 C1g = fy � �(y); y 2 C1fg:
Proof. Since if suÆces to check the claims over an extension K � L, we can assume
that (V; q) = (Alt� (A); pfA) for some c:s: algebra A of dimension 16 with an even
symplectic involution (for example A = M4(K) and �(x) = s2xts�1
2 ). Then by
Proposition 4 of Chapter 10,
C =M2(A); �
a bc d
!=
�(d) ��(b)
��(c) �(a)
!; C0 =
A 00 A
!
94 11. Quadratic Forms of Dimension 6
and we can choose e = (1000); f = (00
01). The space V is embedded in C as the set
V =�
0 �A(x)x 0
!; x 2 Alt� (A)
�:
All the claims of Lemma 2 can now easily be explicitly checked.
Proof of Proposition 1. The algebra A = C0e (with 1A = e) is c:s: of dimension
16 over K. We shall construct an even symplectic involution � on A such that
(V; q) ' (Alt� (A); �pfA). Replacing q by �q, we may assume that q represents 1.
Let v 2 V be such that q(v) = 1 and let P = vA. The standard involution � of C
maps P to P since
�(vce) = �f�(c)v = ��(c)ve = vc0e; for c0 = �v�(c)v;
by Proposition 5 of Chapter 5. Let � : A ! A be de�ned by �(va) = �v�(a). We
have �(ab) = �(b)�(a) and va = �2(va) = ��(v�(a)) = v� 2(a), so � 2 = 1. Therefore
� is an involution of A and by Lemma 2,
V e = v � Alt� (A):
We claim that the type of � is independent of the choice of v. Let w 2 V with
q(w) = 1 and let w = vu; u 2 A. If � 0 is de�ned by �(wa) = �w� 0(a), we get
�(vua) = �vu� 0(a) = �v�(ua) = �v�(a)�(u):
On the other hand w 2 V so �(vu) = �vu and �(vu) = �v�(u). It follows that
u = �(u) and � 0 = iu � . Thus � 0 is of even symplectic type if and only if � is of even
symplectic type. To compute the type of � we may now assume (as in the proof of
Lemma 2) that (V; q) = (Alt�1(A); pfA), with �1 an even symplectic involution of A,
and we can choose v = 1. For this choice, we get � = �1, thus � is of even symplectic
type. We �nally check that the map � : V ! Alt� (A) de�ned by xe = v�(x) is an
isometry: we have similarly
V f = Alt� (A) � vand we de�ne : Alt� (A)! Alt� (A) by the formula ��1(a)f = (a)v. We get
(a)a = (a)vva = q(��1(a))e for all a 2 Alt� (A);
so by Lemma 3 (and the discussion following Lemma 3) of Chapter 10 , = � ��A for
some � 2 K� and q(x)e = �pfA(�(x)) (recall that e = 1A !). Since �(v) = e; � = 1
11. Quadratic Forms of Dimension 6 95
and � is an isometry, as claimed.
In view of the next theorem, the algebra A in Proposition 1 is determined up
to isomorphism by the class of similitudes of (V; q).
Theorem 3. Let A and B be c:s: algebras of dimension 16 over K with pfaÆans
pfA and pfB. Then A ' B if and only if the corresponding pfaÆans are similar.
Proof. Let �A be the involution of A and �B the involution of B. Let � : A ~!B
be an isomorphism of algebras and let �0 = ��A��1. Then � is an isomorphism
(A; �A) ~!(B; �0) of algebras with involutions and we get a pfaÆan pf 0B : Alt�0
(B)!K such that � : (Alt�A(A); pfA)! (Alt�
0
(B); pf 0B) is an isometry. Let �0 = iu Æ �B,so
Alt�0
(B) = uAlt�B(B) = Alt�B(B)u�1:
We claim that the map
u�1� : Alt�A(A)! Alt�B(B)
is a similitude of the corresponding pfaÆans. Let
: Alt�B(B)! Alt�B(B)
be de�ned by (y) = �(�A��1(uy))u. We have (y)y = pfA(�
�1(uy)) 2 K for all
y 2 Alt�B(B). Therefore by Lemma 3 of Chapter 10, there exists � 2 K� such that
� = �B and, as claimed, u�1� is a similitude. Conversely, let
' : (Alt�A(A); pfA)! (Alt�B(B); pfB)
be a similitude. By Proposition 4 of Chapter 10, ' induces a graded isomorphism
M2(A) ~!M2(B), where both matrix algebras have the checker{board grading. In
degree zero, we get A� A ' B �B, hence A ' B.
Example 4. Let A = M2(B); B a quaternion algebra with standard involution
b 7! b. The involution
�1 :
a bc d
!7!
d �b�c a
!
96 11. Quadratic Forms of Dimension 6
of A is the tensor product of the two standard involutions. We get
Alt�1(A) =�
b xy �b
!; b 2 B; x; y 2 K
�and for the corresponding pfaÆan (use a splitting !):
pf1
b xy �b
!= �xy � bb
so that (Alt�1(A); pf1) ' H(K) ? (B;�nB). On the other hand, (see Example 5 of
Chapter 10)
�2 :
a bc d
!7!
a cb d
!is of even symplectic type. We get
Alt�2(A) =�
x bb y
!; b 2 B; x; y 2 K
�and for the corresponding pfaÆan:
pf2
x bb y
!= xy � bb
so that (Alt�2(A); pf2) ' H(K) ? (B;�nB) and both pfaÆans are indeed similar.
Corollary 5. There exists a bijection between similitude classes of quadratic spaces
of dimension 6 with trivial Arf invariant and isomorphism classes of c:s: algebras
of dimension 16, with even symplectic involutions. The correspondence is given by
(V; q) 7! A = C0e, where e is a nontrivial idempotent of the centre of C0. Further
1) A is a division algebra if, and only if, (V; q) is anisotropic.
2) A 'M2(D); D a quaternion division algebra if, and only, if (V; q) has Witt
index 1.
3) A ' M4(K) if, and only if, (V; q) ' H(K3).
The class of A in Br(K) is the Witt invariant of (V; q).
Proof. The �rst claim follows from Lemma 10 of Chapter 5, Proposition 1 and
Theorem 3; the others, then from Example 4.
We now apply Theorem 3 to prove a classical result of Albert. Observe that,
again, we do not assume char K 6= 2.
11. Quadratic Forms of Dimension 6 97
Theorem 6. Any c:s: K{algebra of dimension 16, whose class in Br(K) is of order
2, is isomorphic to a tensor product of quaternion algebras.
Proof. Let A be c:s: of dimension 16 such that 2[A] = 0 in Br(K). By Theorem 9 of
Chapter 3, there exists an even symplectic involution � onA. Let pfA : Alt�(A)! K
be the corresponding pfaÆan. We have 1 2 Alt�(A) and pfA(1) = 1. We denote by
b : Alt�(A)� Alt�(A)! K the polar of pfA. Since pfA is nonsingular, there exists
z 62 K � 1 such that b(1; z) = 1. The element z is a root of its pfaÆan characteristic
polynomial, so
z2 = z � r; where r = pfA(z);
and z generates a quadratic separable K{algebra Z. With the notations of Chapter
1, we get [1; z] ' (Z; n0) and
(Alt�(A); pfA) ' [1; z] ? [a; b] ? [c; d]
for some quadratic spaces [a; b]; [c; d]. We remark that [a; b] ? [c; d] has Arf invari-
ant Z (compute the discriminant if char K 6= 2 and the classical Arf invariant if char
K = 2). Let now (V1; q1) = [a; b]; (V2; q2) = [c; d], (V; q) = (V1; q1) ? (V2; q2); and
let eC = C(�q) = A1 A2 with A1 = C(�Æ(q2)q1); A2 = C(�q2) (see Lemma 8 of
Chapter 5). The canonical involution �0 of eC is of even symplectic type by Example
1 of Chapter 9 and C0 has centre Z.
Lemma 7. (Alt�0
( eC); pfeC) = (Z; n0) ? (V; q).
Proof. If suÆces to prove Lemma 7 over some �eld extension K � L so that we
can assume that (V;�q) = (A; n), where A is a quaternion algebra with reduced
norm n. By Example 1 of Chapter 9, we get
Alt�0
( eC) = � x bb y
!; x; y 2 K; b 2 A
��M2(A) = C
and pfeC( xb by) = xy � bb as claimed.
By Lemma 7 A and eC = A1 A2 have isometric pfaÆans and by Theorem 3,
A ' A1 A2. This proves Theorem 6. Lemma 7 may also be used to describe the
98 11. Quadratic Forms of Dimension 6
structure of the Cli�ord algebra of a quadratic space of dimension 4 :
Proposition 8. Let (V; q) be a quadratic space of dimension 4. Then
1) C(q) is a division algebra if, and only, if the space (Z(q); n0) ? (V;�q) isanisotropic.
2) C(q) ' M2(D); D a quaternion division algebra if, and only, if the space
(Z(q); n0) ? (V;�q) has Witt index 1.
3) C(q) 'M4(K) if, and only, if (Z(q); n0) ? (V;�q) ' H(K3):
Proof. By Lemma 7 we have (Alt�0
(C(q)); pfC(q)) ' (Z; n0) ? (V;�q) so that
Proposition 8 follows from Corollary 5.
Example 9. Let A = A1 A2 be a tensor product of quaternion algebras and
let � = �1 �2 be the tensor product of the corresponding standard involutions.
Assume that char K 6= 2 and let A0i = (K � 1)? � Ai for the reduced norms. Then
Alt�(A) = A01 1 + 1 A0
2 ' A01 � A0
2:
We claim that
pfA(x1 + x2) = nA1(x1)� nA2(x2) for xi 2 A0i;
so
(Alt�(A); pfA) ' (A01; nA1) ? (A0
2;�nA2):
It suÆces to check the formula over a �eld extension K � L. Hence we can assume
that A1 =M2(K). By Example 4, we have
Alt�(A) =�
b xy �b
!; x; y 2 K; b 2 A2
�:
Writing b as z + b0; b0 2 A02, we get b = z � b0,
b xy �b
!=
z xy �z
!+
b0 00 b0
!and
pfA
b xy �b
!= �xy � z2 � n(b0) = det
z xy �z
!� n(b0);
as claimed. The quadratic space (A01; nA1) ? (A0
2 � nA2) is called the Albert form
of A1 A2. We denote it by Q(A1; A2). In view of Theorem 3, two tensor prod-
ucts A1 A2 and B1 B2 of quaternion algebras are isomorphic if and only if the
11. Quadratic Forms of Dimension 6 99
Albert forms Q(A1; A2) and Q(B1; B2) are similar (assuming char K 6= 2). This
result is due to Jacobson. We mention two more results of Albert which are easy
consequences of Theorem 3. We assume char K 6= 2.
1) Let A; B; C be quaternion algebras. If [A][B][C] = 0 in Br(K), then
Q(A;B) is isotropic.
2) Let A; B be quaternion algebras. Then A B ' M2(C) if, and only if,
there exists a quadratic separable algebra Z which is contained in A and B.
1) is clear by Theorem 3. We sketch a proof of 2). If Z � A and Z � B then
(A; nA) ' (Z; n0) ? (V1; q1) and (B; nB) ' (Z; n0) ? (V2; q2) so that (A0; nA)
and (B0; nB) represent a (nonzero) common element. If follows that Q(A;B) is
isotropic. Conversely if AB 'M2(C); Q(A;B) is isotropic and (A0; nA); (B0; nB)
represent a common element � 2 K. If � = 0 A ' M2(K); B ' M2(K) and
the claim is obvious (take Z = K � K). If � 6= 0 and q(v1) = � = q(v2); then
Z = K � 1�K � v1 ' K � 1�K � v2 is the algebra Z.
We now consider quadratic spaces (V; q) of dimension 6 with arbitrary Arf
invariant. Let Z = Z(q) be the centre of the even Cli�ord algebra C0 of (V; q). Let
� be the standard involution of C = C(q) and let
Alt�(C1) = fx� �(x); x 2 C1g:
Lemma 10. The inclusion V ! C1 induces an isomorphism
Z V ~!Alt�(C1):
Proof. Since it suÆces to prove Lemma 10 over a �eld extension K � L, we may
assume that (V; q) = (Alt� (A); pfA) for A a c:s: K{algebra of dimension 16 with an
even symplectic involution � : a 7! a. The claim then follows easily from Example 4.
We now assume that q represents 1, let v 2 V with q(v) = 1, so v2 = 1 in C.
We have �(v) = �v and �1 = iv Æ � is a Z{linear involution of C0 (observe that �
is �0{semilinear, where �0 is the standard involution of Z). We claim that �1 is an
involution of C0 of even symplectic type. The type does not depend on the choice
of v 2 V (such that v2 = 1), so we can take (V; q) = (Alt� (A); pfA) as above and
100 11. Quadratic Forms of Dimension 6
v = (0110) = 1A. Then
�1
a 00 b
!=
�(a) 00 �(b)
!:
Since � is of even symplectic type, �1 is of even symplectic type. By de�nition of
�1, the maps C1 ! C0 given by �v : x! vx and �v : x 7! xv induce isomorphisms
�v; �v : Alt�(C1) ~!Alt�1(C0):
Let pf1 : Alt�1(C0) ! K be the pfaÆan given by �1 and let �1 : Alt�1(C0) !Alt�1(C0) be such that �1(x)x = x�1(x) = pf1(x) for all x 2 Alt�1(C0). We claim
that �v is an isometry
�v : (Z V; 1 q)! (Alt�1(C0); pf1):
We have (1 q)(x) = �x�(x) for all x 2 Z V = Alt�(C1). Let : Alt�1(C0) !Alt�1(C0) be de�ned by (xv) = �v�(x) for x 2 ZV . We get xv (xv) = �x�(x) =(1q)(x), so = ��1 for some � 2 K�. Putting x = v shows that � = 1; = �1 and
as claimed, �v is an isometry. Further we know that V = fx 2 Z V j �(x) = �xgor, since �(x) = �v�1(xv),
V = fx 2 Z V j v�1(xv) = xg:(11.1)
We are now ready to prove the following
Theorem 11. Let (V; q); (V 0; q0) be quadratic spaces of dimension 6 and let
C0 = C0(q); C00 = C0(q
0) with the standard involutions. Then (V; q) and (V 0; q0) are
similar if and only if C0 and C00 are isomorphic as algebras{with{involution.
Proof. We assume that the centre Z of C0 is a �eld (the other case is similar).
Since ' maps Z to the centre Z 0 of C 00, we view C 0
0 as a Z{algebra through '. Then
' is a Z{linear isomorphism. Let ' : C0 ~!C 00 be an isomorphism as algebras{with{
involution, i.e. such that '� = �'; � the standard involution. We can assume
that V and V 0 represent 1 (since only similitude classes matter), so let v 2 V
with q(v) = 1 and v0 2 V 0 with q0(v0) = 1. Let �1 be the involution iv Æ � of C0
and �01 = iv0 Æ �. Further let pf1 : Alt�1(C0) ! Z, pf 01 : Alt�0
1(C 00) ! Z be the
corresponding pfaÆans and
�1 : Alt�1(C0)! Alt�1(C0); �01 : Alt
�01(C 00)! Alt�
0
1(C 00)
11. Quadratic Forms of Dimension 6 101
be such that x�1(x) = pf1(x); x0�01(x0) = pf 01(x). The involution �
0 = '�1 '�1 of
C 00 is of even symplectic type and, writing �0 = iu Æ �01; u 2 C�
0 , we get �01(u) = u
or v0�(u)v0 = u. Since '� = � ', it follows that ' iv � '�1 = ' iv '
�1 � = iu iv0 �
so that ' Æ iv = iuv0 Æ ' or '(vxv) = uv0'(x)v0u�1. Replacing x by v x v, we get
'(x) = uv0'(vxv)v0u�1, so there is � 2 Z� such that �uv0 = v0u�1. This, together
with v0�(u)v0 = u, implies that �(u)u 2 Z�, so also u�(u) 2 Z�. The Z{linear
isomorphism C0 ! C 00; x 7! '(x)u maps Alt�1(C0) to Alt�
0
1(C 00) since
'(Alt�1(C0)) = Alt�0
(C 00) = Alt�
0
1(C 00) � u�1
= u � Alt�1(C 00):
Let now : Alt�1(C0)! Alt�1(C0) be de�ned by
(x) = '�1(u�01('(x)u)):
Since
x (x) = '�1('(x)u�01('(x)u)) = pf 01('(x)u) 2 Z;there exists � 2 Z� such that = ��1 and x 7! '(x)u is a similitude
(Alt�1(C0); pf1)! (Alt�0
1(C 00); pf
01)
with multiplier �. We compute �. By the discussion preceeding Theorem 11, the
map x 7! '(xv)uv0 is a similitude Z (V; q)! Z (V 0; q0) with multiplier �. Since
(1 q0)(x0) = �x0�(x0), we get (1 q0)('(xv)uv0) =
('(xv)uv0)�('(xv)uv0) = '(xv)u�(u)�'(xv). Now u�(u) 2 Z and �' = '�, so
(1 q0)('(xv)uv0) = u�(u)(1 q)(x). It follows that � = u�(u) and �(�) = �0(�) =
�, so � 2 K�. We claim that, in fact, x 7! '(xv)uv0 restricts to a similitude
(V; q)! (V 0; q0) with multiplier �. We have by (11.1)
V = fx 2 Z V j v�1(xv) = xg; V 0 = fx0 2 Z V 0 j v0�01(x0v0) = x0g
so we have to check that v0�01('(xv)u) = '(xv)uv0 if x = v�1(xv): By de�nition of
and the fact that � = u�(u), we get
�(u)'�1(xv) = �01('(xv)u) for xv 2 Alt�1(C0):
On the other hand we obtain, using that x = v�1(xv),
�(u)'�1(xv) = �(u)'(vx) = �(u)'(v(xv)v)
= �(u)uv0'(xv)v0u�1 = v0'(xv)u�(u)�(u)�1v0 = v0'(xv)uv0:
102 11. Quadratic Forms of Dimension 6
Thus
v0'(xv)uv0 = �01('(xv)u) or '(xv)uv0 = v0�01('(xv)u)
as claimed.
Remark 12. The �rst part of Corollary 5 is a consequence of Theorem 11 if
Z = K � K. In fact the algebra C0 is isomorphic to a product A � Aop and the
standard involution � corresponds to the involution �A : (a; b0) 7! (b; a0). Any iso-
morphism of algebras A ! B extends to an isomorphism A � Aop ! B � Bop of
algebras{with{involution. (Observe that the standard involution � of C0 is of the
second kind).
Remark 13. Assume that the Arf invariant of (V; q) is not trivial, i.e. Z = Z(q) is
a �eld. Since
Z (V; q) ' (Alt�1(C0); �pf1)
for some � 2 Z� and some even symplectic involution �1 of C0, it follows from
Corollary 5 that
1) C0 is a division algebra if, and only, if Z (V; q) is anisotropic.
2) C0 ' M2(D); D a quaternion division algebra over L if, and only, if
Z (V; q) has Witt index 1.
3) C0 'M4(Z) if and only if Z (V; q) ' H(Z3).
Remark 14. Let B be a c:s: Z{algebra, Z a separable quadratic �eld extension
of K. The corestriction cor(B) is a c:s: K{algebra and B 7! cor(B) induces a
homomorphism cor: Br(Z) ! Br(K). As shown by Arason for �elds of character-
istic di�erent from 2 or (in a more general context, in particular for �elds of any
characteristic) by Knus{Parimala{Srinivas, the sequence
2Br(K)!2 Br(Z)cor!2 Br(K)
is exact. Here 2Br(K) denotes the subgroup of Br(K) of elements of order 2. Let
now C0 be the even Cli�ord algebra of a quadratic space of dimension 6. Since
the standard involution � is of the second kind, the corestriction of C0 is trivial
(Theorem 11 of Chapter 3) and by the exactness of the above sequence, C0 is of the
form Z B for some c:s: K{algebra B. This is also true for forms of dimension
11. Quadratic Forms of Dimension 6 103
4 (see Chapter 9) and dimension 2. We believe that the even Cli�ord algebra of a
quadratic space of arbitrary even dimension is of the form C0 = Z(q) B. This is
clear if n � 2 (4) by Proposition 6 of Chapter 5 and if CharK 6= 2 by Lemma 8 of
Chapter 5. Observe that the algebra B is not uniquely determined by the form q
(except if Z(q) = K �K).
Let A be a c:s: Z{algebra of dimension 16 which is isomorphic to the even Cli�ord
algebra C0 of a quadratic space (V; q) of dimension 6 with Arf invariant Z. As we
have seen a necessary condition for such an A is that A has an involution b 7! b�
of the second kind (necessary and suÆcient conditions are given in Knus{Parimala{
Sridharan). Let � be another involution of A of the second kind. By Lemma 12
of Chapter 3, � = �g for g 2 A� such that g� = g (�g denotes the involution
�g(x) = gx�g�1). Further (A; �g) ' (A; �g0) , g0 = �cgc� for � 2 K� and c 2 A�.
In this case we call g and g0 similar. By Theorem 10, the similitude classe of (V; q)
corresponds to the similitude class of g.
Remark 15. Let (V; q) and (V 0; q0) be quadratic spaces of dimension 6 with the
same Arf invariant. We choose as a representative of the Arf invariant the centre
Z of C0(q) and view C0(q0) as a Z-algebra through the choice of an isomorphism
Z ~!Z(C0(q0)). We claim thatC0(q) and C0(q
0) are isomorphicK{algebras if and only
if the quadratic spaces (V; q)Z and (V 0; q0)Z are similar. Let ' : C0(q) ~!C0(q0)
be an isomorphism of K-algebras, so that ' 1 is an isomorphism of the even
Cli�ord algebras of the spaces (V; q) Z and (V 0; q0) Z. Since Z Z ~!Z � Z
( an isomorphism being � : x y 7! (xy; x�0(y)), these spaces have trivial Arf
invariant. Hence, by Corollary 5, they are similar. Conversely, if the quadratic
spaces (V; q) Z and (V 0; q0) Z are similar, there exists an isomorphism of Z-
algebras C0(q)Z ~!C0(q0)Z (see Lemma 10 of Chapter 5). The isomorphism � :
ZZ ~!Z�Z composed with the projection Z�Z ! Z onto the �rst factor induces
a homomorphism of K-algebras C0(q0) Z ! C0(q
0). Let ' be the composition
C0(q)! C0(q) Z ~!C0(q0) Z ! C0(q
0):
The map ' is K-linear and maps the centre Z of C0(q) into the centre Z0 of C0(q
0).
By the uniqueness of the standard involution of quadratic algebras, the restriction
'Z of ' to Z is a morphism (Z; nZ) ! (Z 0; nZ0) of quadratic spaces of the same
104 11. Quadratic Forms of Dimension 6
dimension. Since 'Z is not the zero map, 'Z is an isomorphism. If we view C0(q0)
as a Z-algebra through 'Z, ' is a Z-homomorphism between c:s: Z-algebras of the
same dimension. Thus ' must be an isomorphism.
We now compute the special Cli�ord group, the spin group and the Lie algebra
of a quadratic space (V; q) of dimension 6. We may assume that q represents 1, so let
v with q(v) = 1. Let u 2 C�0 . With the same notation as in the discussion preceding
Theorem 10, the condition iu(V ) � V implies uxu�1v 2 Alt�1(C0) for all x 2 V , in
particular �1(uxu�1v) = uxu�1v; so uxu�1v = vv�(u�1)vxv�(u)v and �(u)u must
be in the centre of C0. Since �0(�(u)u) = �(�(u)u) = �(u)u, we get �(u)u 2 K.
To simplify notations we put �(u)u = u�(u) = �(u). Further since by (11.1)
V = fx 2 ZV j v�1(xv) = xg; iu(V ) � V is equivalent with v�1(uxu�1v) = uxu�1
for all x 2 V . We have, using that �1(ay�1(a)) = n(a)�1(a)�1�(y)a�1; n the
reduced norm,
v�1(uxu�1v) = v�1(uxv�(u)
�1�1(u))
= v�(u)�1�1(u)�1�(xv)u1n(u)
= �(u)�1�(u)�1v�(xv)u�1n(u)
= �(u)�2n(u)uxu�1:
Thus iu(V ) � V is equivalent with n(u) = �(u)2 and
S�(q) = fu 2 C�0 j n(u) = �(u)2g
Spin(q) = fu 2 S�(q) j �(u) = 1g:
Let Z be a quadratic �eld extension of K. For any c:s: Z{algebra A with an
involution of the second kind �, we de�ne an involution � 7! �� of Mn(A) by
�� = (�(aij))t if � = (aij) 2Mn(A);
i.e. � 7! �� is the tensor product of the involution � of A with the involution
x 7! xt of Mn(Z). We call a matrix g 2Mn(A) hermitian if g� = g and we say that
two hermitian matrices g; g0 are congruent if there exists x 2 GLn(A) such that
g0 = xgx�.
11. Quadratic Forms of Dimension 6 105
Let g 2Mn(A) be an hermitian matrix. We call the group
GUn(A; g) = fx 2 GLn(A) j xgx� = �g; � 2 Kg
the group of unitary similitudes of g, the group
Un(A; g) = fx 2 GLn(A) j xgx� = gg
the unitary group of g and
SUn(A; g) = fx 2 Un(A; g) j nMn(A)(x) = 1g
the special unitary group of g. With these notations we get
Spin(V; q) = SU1(C0; 1):(11.2)
We consider some special cases. First let
(V; q) ' (Z; nZ) ? (B; �nB);(11.3)
where B is a quaternion algebra over K and � 6= 0 2 K. We have
C(q) ' C(B; �nB) bC(Z; nZ):By Example 1 of Chapter 9, C(B; �nB) is isomorphic to M2(B) with the checker-
board gradation. The standard involution is given by a bc d
!7!
�a ���1�c���b �d
!; a; b; c; d 2 B;
where a 7! �a is the standard involution of B. Further by Example 3 of Chapter 7
C(Z; nZ) = (1; Z=K] ' Z � uZ;
with the multiplication rules xu = u�0(x); u2 = 1 (�0 is the involution of Z) and
the standard involution of C(Z; nZ) corresponds to x+uy 7! �0(x)u�uy; x; y 2 Z.We de�ne a graded isomorphism
: M2(B) b(Z � uZ) ~!M2(B (Z � uZ))
by a bc d
! (x + uy) 7!
a (x + uy) b (x+ uy)c (x� uy) d (x� uy)
!:
106 11. Quadratic Forms of Dimension 6
We use to identify C(q) with M2(B (Z � uZ)). In particular we get
C0 =
B Z B uZB uZ B Z
!
and it is easy to check that the map a x1 b ux2c ux3 d x4
!7! a x1 b �0(x2)c x3 d �0(x4)
!
is an isomorphism ' : C0(q) ~!M2(B Z). Let � be the standard involution of
C0(q). We get
�
a x1 b ux2c ux3 d x4
!=
�a �0(x1) �c ��1ux3��b ux2 �d �0(x4);
!so
'�'�1
a x1 b x2c x3 d x4
!=
�a �0(x1) ��1�c �0(x3)��b �0(x2) �d �0(x4)
!Thus the transport of � to M2(B Z) is the involution
� � Æ
!7! � � Æ
!�=
1 00 �
! ~� ~�
~ ~Æ
!t 1 00 �
!�1
;
where e� = �a �0(x) for � = a x 2 B Z. In view of (11.2) we get
Spin((B; �nB) ? (Z; nZ)) ' SU2(B Z; diag(1; �));(11.4)
If (V; q) is a quadratic space with Arf invariant [Z] such that
(M; q) Z ' ((B; nB) ? (Z; nZ)) Z
then, by Remark 15, C0(q) ' M2(B Z). The standard involution of C0(q) is of
the form �(x) = gx�g�1 for some hermitian matrix g 2M2(B Z) (i.e. g = g�). In
this case
Spin(V; q) ' fx 2 SL2(B Z) j xgx� = gg ' SU2(B; g):
We observe that g is determined up to similarity by (V; q) (see Remark 15). If , in
(11.4), B =M2(K) (and � = 1), we get
Spin(M; q) ' SU4(Z; s2)
since the standard involution ofM2(K) is � 7! s1�ts�1
1 for � 2M2(K). If i 6= 0 2 Zis an element of trace zero, the matrix is1 2 M2(Z) is hermitian and is congruent
to (0110). In fact we have (01
i0)(
01
10)(
0�{
10) = is1. Thus
Spin(H(K2) ? (Z; nZ)) ' SU4(Z; diag(1; 1; �1; �1)):
11. Quadratic Forms of Dimension 6 107
Finally we assume that there exists an isomorphism B Z ~!M2(Z) of algebras{
with{involution where the involution on BZ is the tensor product of the standard
involutions and the involution on M2(Z) is the involution
� =
a bc d
!7! �� =
�0(a) �0(b)�0(c) �0(d)
!t:
Then
Spin((Z; nZ) ? (B; �nB)) ' SU4(Z; diag(1; 1; �; �)):
This is the case for B = IH and Z = IC. In fact the isomorphism IH = IC � j IC 'M2( IC) given by
(x� jy) z 7! x �y�y �x
!z
has the wanted property.
We now assume that (V; q) has trivial Arf invariant and that (V; q) represents 1. By
Proposition 1, we can choose (V; q) = (Alt� (A); pfA) for some c:s: algebra A with an
involution � of even symplectic type. Then C0 ' A � A and �(a; b) = (�(b); �(a)).
It follows that
Spin(V; q) ' f(a; �(a)�1); a 2 SL1(A)g ' SL1(A)
and the map S� : Spin(V; q)! SO(V; q) is given by
a 2 SL1(A) 7! S�(a)(x) = �(a)�1xa�1:
If (V; q) is anisotropic, A is a division algebra. If (V; q) has Witt index 1, (V; q) is
similar toH(K) ? (B; nB), B a quaternion division algebra with standard involution
b 7! b (by Example 4). Then A =M2(B) with the involution
�
a bc d
!=
a cb d
!:
So
Spin(V; q) ' SL2(B):
If (V; q) ' H(K3), we have B =M2(K) in the above formula. Therefore A =M4(K)
with the involution �(x) = s2xts�1
2 and
Spin(V; q) ' SL4(K):
108 11. Quadratic Forms of Dimension 6
By de�nition of pfA, the map � : x 7! s�12 x is an isometry
� : (V; q) = (Alt� (M4(K)); pfM4(K)) ~!(Alt4(K); pf) = H(K3):
Let u 2 Spin(Alt� (M4(K)); pfM4(K)). Since
�S�(u)��1(x) = s�12 (s2u
ts�12 )�1s2xu
�1 = (ut)�1xu�1;
the map S� corresponds through � to the map
SL4(K)! SO(Alt4(K); pf) = SO(H(K3))
de�ned by u 7! (x 7! (ut)�1xu�1).
We now compute the Lie algebra so(q). Let
S(C0; �) = fx 2 C0 j x+ �(x) = 0g:
We obviously have so(q) = [V; V ] � S(C0; �), thus
so(q) = [so(q); so(q)] � S(C0; �)0 = [S(C0; �);S(C0; �)]:
We claim that
so(q) = S(C0; �)0:
It suÆces to verify that DimKS(C0; �)0 = 15. We can assume that (V; q) = (A; pfA)
so
S(C0; �) = f(a;�a); a 2 Ag ' A:
Taking further A = M4(K), we get DimK[A;A] = 15 as claimed. We leave it as an
exercise to compute so(q) for the di�erent cases considered above.
We �nally compute the group GO+(q) of special similitudes of a quadratic space
of dimension 6. Let Z = Z(q) be the centre of C0. Let n0(x) = x�0(x) be the norm
of Z; n the reduced norm of C0 (as Z{algebra), � the standard involution of C and
�(u) = u�(u). Let GU1(C0) = fu 2 C0 j u sigma(u) 2 Z�.
Theorem 16. LetG be the subgroup of Z��GU1(C0) of pairs (z; u) with z�0(z)�1 =
n(u)�(u)�2. There exists an isomorphism
G=f(z2; z�1); z 2 Z�g ~!GO+(q):
11. Quadratic Forms of Dimension 6 109
Proof. Let C = C0 � C1 be the Cli�ord algebra of (V; q). By Lemma 10, we may
identify Z V with Alt�(C1). For (z; u) 2 Z� � GU1(C0), let �(z; u) : Alt�(C1)!
Alt�(C1) be the map �(z; u)(x) = zux�(u). Since (1 q)(x) = �x�(x); �(z; u) is asimilitude of ZV with multiplier n0(z)�(u)
2 2 K�. We claim that �(z; u) restricts
to a similitude of V if and only if z�0(z)�1 = n(u)�(u)�2. By (11.1) and with
the notations introduced after Lemma 10, we have to check that v�1(zux�(u)v) =
zux�(u) for x 2 V: We get
v�1(zux�(u)v) = �0(z)�1(uxv�1(u)):
It now follows from pf1(uc�1(u)) = n(u)pf1(c) for c 2 C0 that
�1(uc�1(u)) = n(u)�1(u)�1�1(c)u
�1;
hence
v�1(zux�(u)v) = �0(z)n(u)�1(u)�1�1(xv)u
�1
= �0(z)n(u)�(u)�2ux�(u);
using that v�1(xv) = x; �1(u) = vuv and �(u) = u�(u). Thus, as claimed,
zux�(u) 2 V if and only if �0(z)n(u)�(u)�2 = z. The multiplier of �(z; u) is
n0(z)�(u)2. The extension of �(z; u) to C0 is given on elements xy; x; y 2 V , by
C0(�(z; u))(xy) = n0(z)�1�(u)�2zux�(u)xuy�(u) = uxyu�1
(see Lemma 10 of Chapter 5), thus �(z; u) 2 GO+(q). If �(z; u) = 1, then, by the
above formula, u 2 Z and zu2 = 1. Therefore � induces an injective map
�0 : G=f(z2; z�1); z 2 Z�g ! GO+(q):
We check that �0 is surjective. Let
E =
�C0 �C1
C1 C0
!� M2(C):
The map Z V = Alt�(C1)! E; x 7! (0xx0) extends to an isomorphism
Z C ~!E of Z{algebras and we identify Z C with E through this isomorphism.
Let f : V ! V be a special similitude with multiplier �. The map 0 xx 0
!2 Z V 7!
0 ��1f(x)
f(x) 0
!2 E = Z C
110 11. Quadratic Forms of Dimension 6
extends to an automorphism ' of ZC. As in the proof of Theorem 12 of Chapter
9, one checks that ' = iu with u 2 (Z C0)�. As in Lemma 4, Chapter 6, we
have �(u)u 2 Z�. The standard involution of Z C viewed as an involution of E
corresponds to a bc d
!7�!
�(d) �(b)�(c) �(a)
!:
Since
Z C0 =
�C0 00 C0
!;
the condition �(u)u 2 Z� implies
u =
��1�(d)�1 0
0 d
!
for some d 2 C�0 and � 2 Z�. If follows that (1 f)(x) = �dx�(d). Since if 1 f
descends to f : C ! V , we must have ��0(�)�1 = n(d)�1�(u)2 and f = �(�; d).
As for Spin(V; q), the computation of GO+(q) given in Theorem 16 can be
applied to a lot of cases (see the Appendix A), depending on the type of (V; q). We
only consider the cases of trivial Arf invariant. We may assume that (V; q) represents
1, so (V; q) = (Alt� (A); pfA) for some c:s: K{algebra of dimension 16 with an even
symplectic involution � . We have C0 ' A� A; �(a; b) = (�(b); �(a)),
GU1(C0) = f(��(d)�1; d); d 2 A�; � 2 K�g
so �(u) = � for u = (��(d)�1; d) and
n(u)�(u)�2 = (�2�0(nA(d))�1; nA(d)�
�2) = z�0(z)�1
for z = (1; nA(d)��2), where nA is the reduced norm of A. Projecting Z� �C�
0 onto
the second factor K� � A�, we get
G ' K� � A�
and
GO+(q) ' K� � A�=f(�2; ��1); � 2 K�g:
Chapter 12
Quadratic Forms of Dimension 5
The theory of 12{regular quadratic forms of dimension 5 is a by{product of the
theory of quadratic spaces of dimension 6. We begin with an example:
Example 1. Let A be a c:s: K{algebra of dimension 16 with an even symplectic
involution �A and let
Alt�A(A) = fx+ �A(x); x 2 Ag
be the corresponding set of alternating elements. Let pfA : Alt�A(A) ! K be the
reduced pfaÆan. For any a 2 Alt�A(A). we have a pfaÆan characteristic polynomial
��(X; a) = X2 � pft(a)X + pfA(a) (see Chapter 10) and ��(a; a) = 0. To simplify
notations, we put
Alt�A(A) = A+; pft(a) = t+(a); pfA(a) = n+(a)
and
A0+ = fa 2 A+ j t+(a) = 0g:
We check that (A0+; n+) is
12{regular of dimension 5 and compute its Cli�ord algebra.
Let Z be the graded quadratic K{algebra with a generator z of degree 1 such that
z2 = 1. We grade A Z by assuming that the elements of A are of degree 0.
Proposition 2. 1) (A0+;�n+) is 1
2{regular of dimension 5. 2) There exists a
graded isomorphism C(A0+;�n+) ~!A Z. In particular C0(A
+;�n+) ~!A and the
12{discriminant of (A0
+;�n+) is trivial. 3) The standard involution of C(A+;�n+)corresponds to the involution �A �0, where �0 is the standard involution of Z, i.e.
�0(z) = �z.
112 12. Quadratic Forms of Dimension 5
Proof. If suÆces to prove 1) over an extension K � L so that we can assume that
A = M2(B); B a quaternion algebra with standard involution b 7! b and �A is
given by (see Example 5 of Chapter 10)
�A
a bc d
!=
a cb d
!:
Then
A+ =�
x bb y
!x; y 2 K; b 2 B
�;
n+
x bb y
!= x y � b b; t+
x bb y
!= x + y;
A0+ =
� x bb �x
!; x 2 K; b 2 B
�;
and (A0+;�n+) ' h1i ? (B; nB) is
12{regular. We have a2 = �n+(a) since a satis�es
its pfaÆan characteristic polynomial and t+(a) = 0. Thus the map
' : A0+ ! A Z; '1(a) = a z;
extends, by the universal property of the Cli�ord algebra, to a homomorphism of
graded K{algebras
' : C(A0+;�n+)! A Z:
Let Z(C) be the centre of C(A0+;�n+). We claim that the restriction of ' to Z(C)
is an isomorphism Z(C) ~!Z. By passing to an extension K � L, we may assume
that (A+;�n+) = h1i ? (M2(K); det). Then Z(C) is generated by
z1 = e0(1� 2e1f1)(1� 2e2f2);
where e0 is the generator of h1i and the elements
e1 =
0 01 0
!; f1 =
0 10 0
!; e2 =
1 00 0
!; f2 =
0 00 1
!
are hyperbolic pairs in M2(K) (see Chapter 5). A computation (!) shows that
'(z1) = z, thus as claimed ' : Z(C) ~!Z. It then follows that ' maps C0(A0+;�n+)
to A and is an isomorphism. To prove 3), we observe that the restriction of �A �0to the image of A0
+ is �1.
Let (V; q) be a 12{regular quadratic form of dimension 5. Since there exists
an extension K � L such that L (V; q) ' h1i ? H(K2) ' h1i ? (M2(K); det),
12. Quadratic Forms of Dimension 5 113
it follows from Proposition 2 that the standard involution of C0(V; q) is of even
symplectic type. Let C0+ be the corresponding set of alternating elements and
n+ : C0+ ! K the reduced pfaÆan. Let z 2 Z(q) be a generator of degree 1
with z2 = s and [s] = 12Æ(q) in K�=K�2 (see Proposition 5 of Chapter 5), and let
� : V ! C0 be the map induced by v 7! zv. By going over to an extension K � L
and applying Proposition 2, we get that � is an isometry
(V; q) ~!(C 00+;�sn+):(12.1)
Theorem 3. 1) Let (V; q) be a 12{regular quadratic form of dimension 5. There ex-
ists � 2 K� and A c:s: of dimension 16 with an even symplectic involution �A such
that (V; q) ' (A0+; �n+). The class of � in K�=K�2 is equal to 1
2Æ(q) and (A; �A) is
unique up to isomorphism of algebras{with{involution. 2) Two 12{regular quadratic
forms (V; q); (V 0; q0) are isometric if and only if 12Æ(q) = 1
2Æ(q0) and C0(q) ' C0(q
0)
as algebras{with{(standard){involution. 3) (V; q); (V 0; q0) are similar if and only if
C0(q) ' C0(q0) as algebras{with{involution.
Proof. 1) We take � = �s and A = C0 with the standard involution.
2) It suÆces to check that an isomorphism ' : (A; �A)! (B; �B) of algebras{with{
involution induces an isometry (A0+; n+) ~!(B0; n+). Clearly ' maps A+ to B+. Let
�A : A+ ! A+ be such that x�A(x) = �A(x)x = pfA(x) and �B : B+ ! B+ such
that x�B(x) = �B(x)x = pfB(x). Let = '�A'�1 : B+ ! B+. Since is a
K{linear isomorphism such that (x)x 2 K for all x 2 B+; = ��B for � 2 K�.
It follows from (1) = 1 that � = 1 and '�A = �B'. Thus ' is an isometry
(A+; n+) ~!(B+; n+). The fact that ' restricts to an isometry (A0+; n+) ~!(B0
+; n+)
follows from the formula x+ �A(x) = t+(x) (see Example 5 of Chapter 10). Finally
3) follows from (12.1).
Remark 4. In view of Proposition 6 of Chapter 7, Corollary 3 of Chapter 8, Theo-
rem 7 of Chapter 9, Theorem 10 of Chapter 11 and Theorem 3 above, two quadratic
forms (V; q); (V 0; q0) which are of dimension � 6 and are nonsingular in even dimen-
sion, resp. 12{regular in odd dimension, are similar if and only if the even Cli�ord
algebras C0(q) and C0(q0) are isomorphic as algebras{with{involution, where the
involutions are the standard involutions.
114 12. Quadratic Forms of Dimension 5
The next result is due to Tamagawa in the case of (signed) trivial discriminant.
We do not know if it holds for 12{regular forms of dimension 5 in characteristic 2.
Proposition 5. Let K be a �eld of characteristic di�erent from 2 and let (V; q)
be a quadratic space of dimension 5 and signed discriminant [d] 2 K�=K�2. The
following conditions are equivalent:
1) (V; q) represents d.
2) (V; q) ' hdi ? (B; �nB) for some quaternion algebra B and � 2 K�.
3) C0(V; q) 'M2(B) for some quaternion algebra B.
Thus C0 is a division algebra if and only if (V; q) does not represent d.
Proof. 1) implies 2) by Remark 11 of Chapter 9, since hdi? is a quadratic space of
dimension 4 with trivial discriminant (and hence trivial Arf invariant). If 2) holds,
then C0(V; q) ' C(B;�d�nB) by Lemma 8 of Chapter 5 and C(B;�d�nB) 'M2(B)
by Example 1 of Chapter 9. Let now C0(V; q) ' M2(B). By Lemma 8 of Chapter
5, we have
C0(h�di ? q) ' C(dq) ' C0(dq) Z(dq) ' C0(q)� C0(q)
(since Z(dq) ' K � K). On the other hand h�di ? q is of dimension 6 and has
trivial Arf invariant. By Corollary 5 of Chapter 11 (applied to the form h�di ? q
and A = C0(dq)), we get that C0(q) ' C0(dq) 'M2(B) if and only if h�di ? q has
Witt index � 1. Thus C0(q) 'M2(B) implies that
h�di ? q ' h�1i ? h1i ? q0 ' h�di ? hdi ? q0
and
q ' hdi ? q0
by Witt cancellation. As claimed, q represents d.
We now compute the spinor group of a 12{regular form of dimension 5. By
Theorem 3 we may assume that (V; q) = (A0+;�n+) for a c:s: K{algebra A of
dimension 16 with an involution � of even symplectic type. In view of Proposition
2 C0(V; q) ' A and the standard involution on C0 is the involution � of A. For any
12. Quadratic Forms of Dimension 5 115
u 2 A, let �(u) = u�(u) and n(u) = nA(u). We have Z(q) = K � 1 � K � z with z
of degree 1 such that z2 = 1. Through the map V ! C0; v 7! zv, we can identify
V ' A0+ � C1 with A
0+ as a subspace of A = C0. Any isometry ' of V extends to
an automorphism of A = C0 as an algebra{with{involution, hence to an isometry
of (A+; n+) (or (A+;�n+)), which restricts to ' on A0+. Thus for any u 2 A�; iu
is an isometry of A0+ (i.e. iu(A
0+) � A0
+) if and only if iu(A+) � A+. By Lemma
4 of Chapter 6, �(u) 2 K� if u 2 S�(A0+;�n+). On the other hand, obviously
iu(A+) � A+ if u�1 = ��(u); � 2 K�. Thus
S�(A0+;�n+) = fu 2 A� j u�(u) 2 Kg:
We recall that a matrix g 2Mn(A) is hermitian (with respect to �) if g = g�, where
g� = (�(aij))t if g = (aij):
The group
GUn(A; g) = fx 2 GLn(A) j xgx� = �g; � 2 K�g
is the group of (unitary) similitudes of g (see Chapter 11). With this de�nition,
S�(A0+;�n+) = GU1(A; 1)
and
SO+(A0+;�n+) ' GU1(A; 1)=K
�:
We claim that
Spin(A0+;�n+) = fx 2 A� j x�(x) = 1g:
We have to check that nA(u) = 1 if u�(u) = 1. By passing to an extension K � L,
we may assume that A = M4(K) and �(a) = s2ats�1
2 . It follows from u�(u) = 1
that s2 = us2ut. Taking pfaÆans shows that det(u) = 1 as claimed. The map
Spin(A0+;�n+)! SO(A0
+;�n+)
is given by u 7! (x 7! uxu�1 = ux�(u)). In fact x 7! ux�(u) is an isometry since
n+(ux�(u)) = nA(u)n+(x) and nA(u) = 1.
We consider now some special cases. Let
(V; q) = h1i ? (B; �nB)
116 12. Quadratic Forms of Dimension 5
for a quaternion algebra B with standard involution b 7! b and � 2 K�. By Lemma
8 of Chapter 5, we have C0(V; q) ' C(B;��nB) and by Example 1 of Chapter 9,
C(B;��nB) 'M2(B) with the standard involution given by
� :
a bc d
!7�!
a ��1c�b d
!=
1 00 �
! a bc d
!t 1 00 �
!�1
:
Therefore
SO+(V; q) ' GU2(B; diag(1; �))=K�
and
Spin(V; q) ' SU2(B; diag(1; �)):
If B =M2(K) and � = 1, then b = s1bts�1
1 , so
S�(h1i ? H(K2)) = fx 2 GL4(K) j xs2xts�12 2 K�g
= GSp4(K)
and
Spin(h1i ? H(K2)) = fx 2 GL4(K) j xs2xts�12 = 1g
= Sp4(K);
if we de�ne the group of symplectic similitudes GSp2m(K) by
GSp2m(K) = fx 2 GL2m(K) j xsmxt = �sm; � 2 K�g:
and the symplectic group Sp2m(K) by
Sp2m(K) = fx 2 GL2m(K) j xsmxt = smg
We �nally show that h1i ? H(K2) can be identi�ed in a natural way with a subspace
of Alt4(K), the set of alternating (4� 4){matrices.
We have
h1i ? H(K2) = (M4(K)0+;�n+)for the involution �(x) = snx
ts�12 , and the map � : x! s�1
2 x is an isometry
� : (M4(K)+;�n+) ~!(Alt4(K);�pf)
by de�nition of n+. Since
M4(K)0+ =�0BBB@
x 0 d �b0 x �c aa b �x 0c d 0 �x
1CCCA 2 M4(K)�;
12. Quadratic Forms of Dimension 5 117
M4(K)0+ has image
�(M4(K)0+) =�0BBB@
0 �x c �ax 0 d �b�c �d 0 xa b �x 0
1CCCA 2 M4(K)�:
We call matrices in �(M4(K)0+) alternating matrices with pfaÆan trace zero and
denote �(M4(K)0+) by Alt04(K). Let u 2 Spin(M4(K)0+;�n+), since � iu ��1(x) =
s�12 us2xu
�1 = ut�1xu�1, the map
s� : Spin(M4(K)0+;�n+)! SO(M4(K)0+;�n+)
corresponds through � to the map
Sp4(K)! SO(Alt04(K);�pf)
de�ned by u 7! (x 7! ut�1xu�1).
We �nally compute the Lie algebra of (V; q) = (A0+;�n+). Let
S(A) = fx 2 A j x+ �(x) = 0g:
We have so(q) � S(A): Since DimKS(A) = 10, we get
so(A0+;�n+) = S(A) = fx 2 A j x+ �(x) = 0g:
Appendix A
A Chart of Results
We summarize in the following chart all the information we have obtained in
these notes on the structure of Cli�ord algebras of forms of dimension � 6. We
restrict to nonsingular forms, so we assume that char K 6= 2 in odd dimensions. We
use the following notations:
` : the dimension of V� : the Witt index of (V; q)L : a separable quadratic �eld extension of KDm : a central division algebra over K of dimension m2
LDm : a central division algebra over L of dimension m2 induced byscalar extension from a central division algebra over K
Mn(R) : the ring of n� n{matrices over R.
` (V; q) Z(q) C0(q) C(q)2 � = 0; 1 2 q(V ) L L M2(K)
� = 0; 1 62 q(V ) L L D2
� = 1 K �K K �K M2(K)
3 � = 0 K �K D2 D2 �D2
� = 0; �(L q) = 0 L D2 LD2
� = 0; �(L q) = 1 L D2 M2(L)� = 1 K �K M2(K) M2(K)�M2(K)� = 1 L M2(K) M2(L)
A. A Chart of Results 119
` (V; q) Z(q) C0(q) C(q)4 � = 0 K �K D2 �D2 M2(D2)
�((V;�q) ? (L; n)) = 0 L LD2 D4
� = 0; �((V;�q) ? (L; n)) = 1 L LD2 M2(D2)� = 1; 1 2 q(H(K)?) L M2(L) M4(K)� = 1; 1 62 q(H(K)?) L M2(L) M2(D2)� = 2 K �K M2(K)�M2(K) M4(K)
` (V; q) Z(q) C0(q) C(q)
5 Let L = K(pd); d 62 K�2 :
� = 0; 1 62 q(V ) K �K D4 D4 �D4
� = 0; 1 2 q(V ) K �K M2(D2) M2(D2)�M2(D2)� = 0; d 62 q(V ); 1 62 q(L V ) L D4 LD4
� = 0; d 62 q(V ); 1 2 q(L V );�(h1i?) = 0 in L V L D4 M2(LD2)� = 0; d 62 q(V ); 1 2 q(L V );�(h1i?) = 2 in L V L D4 M4(L)� = 0; d 2 q(V );�(L hdi?) = 0 L M2(D2) M2(LD2)� = 0; d 2 q(V );�(L hdi?) = 2 L M2(D2) M4(L)� = 1 K �K M2(D2) M2(D2)�M2(D2)� = 1; �(L q) = 1 L M2(D2) M2(LD2)� = 1; �(L q) = 2 L M2(D2) M4(L)� = 2 K �K M4(K) M4(K)�M4(K)� = 2 L M4(K) M4(L)
` (V; q) Z(q) C0(q) C(q)6 � = 0 K �K D4 �D4 M2(D4)
� = 0; �(L q) = 0 L LD4 ?� = 0; �(L q) = 1 L M2(LD2) ?� = 0; �(L q) = 3 L M4(L) ?� = 1 K �K M2(D2)�M2(D2) M4(D2)� = 1; �((V;�q) ? (L; n)) = 1 L M2(LD2) M2(D4)� = 1; �((V;�q) ? (L; n)) = 2 L M2(LD2) M4(D2)� = 2; 1 2 q(H(K2)?) L M4(L) M8(K)� = 2; 1 62 q(H(K2)?) L M4(L) M4(D2)� = 3 K �K M4(K)�M4(K) M8(K)
It would be interesting to know the structure of the Cli�ord algebra C(q) for ` =
6; � = 0 and Z(q) ' L. In particular one would like to know under which conditions
(on q) C(q) is a division algebra. We observe that the case � = 0; �(L q) = 0
cannot occur over IQ : Let d be the signed discriminant of q, so that L = IQ(pd).
By Meyer's theorem (see Serre) any quadratic space of dimension � 5 over IQ, which
is inde�nite over IR, is isotropic. So d must be negative. Let now x 6= 0 2 V .
The quadratic form qjx? ? dqjIQx is isotropic, again by Meyer's theorem, since it is
120 A. A Chart of Results
inde�nite over IR. Thus there exists y ? x such that q(y) = �dq(x) and L q is
isotropic. The case � = 0; �(L q) = 1 cannot occur over IR but can occur over
IQ. An example is given by the form h1; 1; 1; 1; 1; 7i. Let L = IQ(p�7). If L q is
hyperbolic, then Lh1; 1; 1; 1i is hyperbolic, since Lh1; 7i is hyperbolic. Thus thequaternion algebra (�1; �1
L) is a matrix algebra. This is the case if h1; 1i represents
�1 over L. It would follow that �1 is a square modulo 7.
Appendix B
Spinors
We de�ne spinors as in the book of Chevalley and compute some examples from
physics. We shall only consider nonsingular forms and assume that the character-
istic is not equal to 2. More on applications of Cli�ord algebras to physics can be
read for example in the book of Hermann.
I. Spaces of even dimension.
In view of Theorem 8 of Chapter 4, the Cli�ord algebra C = C(q) of a quadratic
space (V; q) of even dimension is simple. Thus there exists up to isomorphism only
one type of simple C(q){modules or, in the language of representation theory, all
irreducible representations of C(q) are equivalent. If we select such a representation,
� : C(q)! EndK(W ), we call W the space of spinors of (V; q), � the spin represen-
tation of C(q) and the restriction �0 of � to C0 = C0(q) the spin representation of C0
. Thus spinors are elements of a �xed simple C(q){module. The representation �
induces representations of �(q); S�(q) and Spin(q). The algebra C0 is either simple
(if its center is a �eld) or the sum of two simple algebras (if Z(q) ' K � K), see
Theorem 8 of Chapter 4.
Lemma 1. Let � : C ! EndK(W ) be an irreducible representation of C. The
restriction �0 of � to C0 is irreducible if C0 is simple and is the sum of two nonequiv-
alent irreducible representations �+0 : C0 ! EndK(W+), ��0 : C0 ! EndK(W
�) if
C0 is not simple.
122 B. Spinors
Proof. Since all irreducible representations of C are equivalent, we may assume
that W is a minimal left ideal of C and that � is the regular representation, i.e.
�(u)y = uy. Let W 0 6= f0g be a subspace of W of minimal dimension which is
invariant under all operations �0(x); x 2 C0, so that the induced representation
C0 ! EndK(W0) is irreducible. Let x 2 V be anisotropic and let W 00 = �(x)W 0.
Since C1 = xC0 = C0x and C0 = xC1 = C1x, we get �0(C0)W00 = W 00 and
�(C)(W 0 +W 00) = W 0 +W 00. Since � is irreducible, we must have W = W 0 +W 00.
If W 0 \W 00 6= f0g, the minimality of W 0 implies that W 0 \W 00 =W 0. Since W 0 and
W 00 have the same dimension, it follows that W = W 0 = W 00. If W 0 \W 00 = f0g,then W = W 0 �W 00. Thus the representation �0 is either irreducible or the sum of
two irreducible representations. Assume now that C0 is not simple. Then C0 has
two types of nonequivalent irreducible representations (see Theorem 8 of Chapter
4). The representation �0 being faithful, these two types occur in �0. Therefore, as
claimed, �0 is the sum of two nonequivalent irreducible representations W+ = W 0
and W� =W 00. This case occurs if K is algebraically closed. Thus we see, by going
over to the algebraic closure, that, if �0 is the sum of two irreducible representations,
these representations cannot be equivalent. It follows that �0 must be irreducible if
C0 is simple.
The elements of W+; W� are called 12{spinors (if W 0 is not equivalent to W 00)
and the induced representations of C0 are the12{spin representations.
Lemma 2. Let (V; q) be a nonsingular quadratic space (of arbitrary �nite di-
mension) non isometric to H(IF3). Let v 2 V be anisotropic and let M = fx 2V j q(x) = q(v)g. 1) The set M generates as a linear space V . 2) C0(q) is generated
as a K{algebra by all products xy; x; y 2 M . 3) S�(q) is a set of generators of
C0(q). 4) Spin(q) is a set of generators of C0(q).
Proof. 1) Let U be the linear hull of M . Let w 2 V be an anisotropic element and
let �w be the re ection at w. Since �w(v) = v � b(v;w)q(w)
w, we get that either w 2 U
or w 2 U?. Since V has an orthogonal basis we must have V = U ? U?. Assume
now that U? 6= 0. Let v 2 M and w 6= 0 2 M?. For any � 6= 0 2 K, the element
�v + w is isotropic, thus �2q(v) + q(w) = q(v) + q(w) = 0 and �2 = 1. There exists
B. Spinors 123
only one �eld K of characteristic not 2 such that �2 = 1 for all � 6= 0 2 K, namely
IF3. Since V = U ? U?, U and U? are nonsingular. It is then easy to see that
DimKU = DimKU? = 1 and V = H(IF3). 2) and 3) are immediate consequences of
1). We check 4). Since C0(q) is generated by all products of two elements of M , it
is also generated by the products a�1xy with a = q(x) = q(y). But a�1xy belongs
to Spin(q).
Proposition 3. Assume that (V; q) is nonsingular of even dimension and (V; q) 6'H(IF3). The spin representations of S�(q) and Spin(q) are either irreducible or the
sum of two irreducible nonequivalent representations.
Proof. Proposition 3 is an immediate consequence of Lemma 1 and Lemma 2.
II. Spaces of odd dimension.
Since we restrict to nonsingular forms, we must have charK 6= 2. The algebra
C0(q) is c:s:, hence its irreducible representations are all equivalent. We �x one,
which we call the spin representation:
�0 : C0(q)! EndK(W ):
The elements of W are called spinors of (V; q).
Proposition 4. The induced representations �0 : S�(q)! EndK(W );
�0 : Spin(q)! EndK(W ) are irreducible.
Proof. As Proposition 3.
If C(q) is simple, then C(q) = L C0(q); K � L a quadratic �eld extension
and �0 can be extended in an unique way to an irreducible representation of C(q):
1 � : L C0(q) = C(q)! EndL(LW ):
We call this representation the spin representation of C(q).
124 B. Spinors
Proposition 5. If C(q) is not simple, then �0 can be extended to exactly 2 non
equivalent irreducible representations of C.
Proof. If C(q) is not simple, we have Z(q) ' K[x]=(x2 � 1). Let z be a generator
of degree 1 of Z(q) such that z2 = 1. Any u 2 C(q) has a unique representation
u = u1 + u2z; ui 2 C0(q). Since z is in the centre of C(q) the maps
'0 : u 7! u1 + u2 and '00 : u1 � u2
are homomorphisms C ! C0 of K{algebras. The representations �0 = �0 Æ '0 and�00 = �0 Æ '00 extend �0 to C. Conversely let � : C ! EndK(W ) be an irreducible
representation which extends �0. Let = �(z) 2 EndK(W ). We have 2 = 1 and
commutes with all elements �0(C0). Therefore W =W1 �W2, where
W1 = fx 2 W j (x) = xg and W2 = fx 2 W j (x) = �xg:
These spaces are invariant under �0(C0). Since � is irreducible, one is zero and the
other W . Thus = �1 and � is one of the representations �0 or �00.
If C(q) is not simple, the two non equivalent representations of C(q) given by
Proposition 5 are called the spin representations of C(q).
In chapter 6 we identi�ed the Lie algebra so(q) with a subalgebra of C0(q). Thus
a spin representation of C0(q) induces, by restriction, a representation of so(q). We
call this representation the spin representation of so(q).
The chart of Appendix A gives an (abstract) description of spin representations
in low dimensions. To have a concrete description, we need explicit representations
of the Cli�ord algebras, i.e. explicitly given simple C(q){modules. We now construct
such representations for three cases occuring in physics. In some of the examples,
we get representations over the real quaternions IH. Writing any quaternion z as
x + jy; x; y 2 IC, we de�ne an injective map � : IH ! M2(IC) through the regular
representation `a(z) = az. In the following we shall identify IH with its image
�(IH) =�
x �yy x
!; x; y 2 IC
��M2(IC):
B. Spinors 125
In particular we get for the pure quaternions
i =
i 00 �i
!j =
0 �11 0
!and k = ij =
0 �i�i 0
!:
Example 6. Let (V; q) = h1; 1; 1i over IR. The algebra C(q) is the Pauli algebra.
We have (see Appendix A)
C(q) 'M2(IC) and C0(q) ' IH:
Explicit isomorphisms can be constructed using the Pauli matrices:
�1 =
0 11 0
!�2 =
0 �ii 0
!�3 =
1 00 �1
!in M2(IC):
Let fe1; e2; e3g be an orthonormal basis of V . We de�ne
'0 : V !M2(IC)
by ei 7! �i; i = 1; 2; 3. By the universal property of the Cli�ord algebra, '0 extends
to a homomorphism ' : C(q) ! M2(IC). It is easy to check that '0 is a IC{linear
isomorphism if we identify IC with Z(q) through i 7! e1e2e3. The subalgebra C0(q)
is mapped by ' onto the IR{subalgebra of M2(IC) with basis 1 00 1
!; �1�2 =
i 00 �i
!; �1�3 =
0 �11 0
!; �2�3 =
0 ii 0
!:
Thus ' restricts to an isomorphism C0 ~!IH. The isomorphism ' : C ~!M2(IC) is a
spin representation. We construct an explicit representation: let e = 12(1 + e3) 2
C(q). We have '(e) = (1000), so '(C(q)e) = (ICIC
00) and we can choose W = (ICIC
00).
Spinors are pairs (xy) 2 IC2. Since Spin(V; q) = fa 2 C0 j nC0(a) = 1g, ' induces an
isomorphism
Spin(V; q) ~! fa 2 IH j nIH(a) = 1g= SU2(IC) =
� x �yy x
!; x; y 2 IC j det
x �yy x
!= 1
�:
The Lie algebra so(q) = [V; V ] is generated by �1�2; �2�3 and �1�3. Let
su2(IC) =�
x �yy x
!; x; y 2 IC j tr
x �yy x
!= 0
�;
we get '(so(q)) = su2(IC).
126 B. Spinors
Example 7. Let (V; q) = h1;�1;�1;�1i over IR. The algebraC(q) is theMinkowski
algebra. By Appendix A, we have
C(q) 'M2(IH) and C0(q) 'M2(IC):
We construct explicit isomorphisms. Let fe0; : : : ; e3g be an orthogonal basis of V
such that q(e0) = 1; q(ei) = �1 i = 1; 2; 3. Let q1 = h1;�1i and q2 = h�1;�1i, so
C(q1) ~!M2(IR) with e0 7!
0 11 0
!and e1 7!
0 1
�1 0
!;
C(q2) ~!IH with e2 7! i; e3 7! j
(where i; j; k = ij are the pure quaternions in IH) and
: C(q) ~!M2(IR) bIH ~!M2(IR) IH =M2(IH)
(the second isomorphism is as in Lemma 3 of Chapter 4). Explicitely
(e0) =
0 11 0
!; (e1) =
0 1�1 0
!;
(e2) =
i 00 �i
!and (e3) =
j 00 �j
!:
It is not obvious that restricts to an isomorphism C0(q) ~!M2(IC). Using a trick
as in the proof of Theorem 8 of Chapter 4, we replace by ' = iu Æ , whereu = (10
0i ) 2M2(IC). We get
'(e0) =
0 �ii 0
!; '(e1) =
0 �i�i 0
!;
'(e2) =
i 00 �i
!and '(e3) =
j 00 j
!:
Then
'(e0e1) =
�1 00 1
!; '(e0e2) =
0 �1
�1 0
!; '(e0e3) =
0 �kk 0
!
'(e1e2) =
0 �11 0
!; '(e1e3) =
0 kk 0
!; '(e2e3) =
k 00 �k
!and
'(e0e1e2e3) =
�k 00 �k
!;
so ' maps C0(q) toM2(IC), with IC = IR(k). The element e = 1+e1e02
is an idempotent
of C(q) such that '(e) = (1000). So C(q)e is a space of spinors. We have
'(C(q)e) =W =
IH 0IH 0
!:
B. Spinors 127
By the general theory of Chapter 9 ' induces an isomorphism Spin(q) ' SL2(IC).
Further the image of so(q) is
sl2(IC) = fx 2M2(IC) j tr(x) = 0g:
Example 8. Let (V; q) = h1; 1; 1;�1i over IR. The algebra C(q) is called the
Majorana algebra. By our chart
C(q) 'M4(IR) and C0 'M2(IC):
We construct explicit isomorphisms. We decompose q as
q = q1 ? q2 with q1 = h�1; 1i and q2 = h1; 1i
and take corresponding orthogonal bases fe0; e1g; fe2; e3g. We have
C(q1) ~!M2(IR); e0 7!
0 1�1 0
!; e1 7!
0 11 0
!
and
C(q2) ~!M2(IR); e2 7!
0 11 0
!; e3 7!
1 00 �1
!:
These two isomorphisms induce
: C(q) ~! C(q1) bC(q2) 'M2(IR) bM2(IR)
~! M2(IR)M2(IR) =M4(IR);
applying Lemma 3 of Chapter 4. Here again it is not obvious that maps C0(q) to
M2(IC) and we �rst modify to get a better representation (as in Example 7). To
simplify notation, we use the symbols
" =
0 11 0
!; � =
1 00 �1
!; � =
0 1
�1 0
!:
Let now ' = iu Æ , where u = (100") 2 M4(IR) = M2(M2(IR)). We have by
de�nition of
(e0) =
0 1
�1 0
!; (e1) =
0 11 0
!;
(e2) =
" 00 �"
!; (e3) =
� 00 ��
!;
where 1 stands for the (2� 2)-unit matrix. So
'(e0) =
0 "
�" 0
!; '(e1) =
0 "" 0
!;
128 B. Spinors
'(e2) =
" 00 �"
!; '(e3) =
� 00 �
!:
If we identify IC with the subalgebra IR("�) of M2(IR) by putting i = "�, we get
'(e0e1) =
1 00 �1
!; '(e0e2) =
0 �1
�1 0
!; '(e0e3) =
0 i�i 0
!;
'(e1e2) =
0 �11 0
!; '(e1e3) =
0 ii 0
!; '(e2e3) =
i 00 �i
!
and
'(e0e1e2e3) =
i 00 i
!
so ' : C0(q) 'M2(IC) as claimed. The element ~e = 1+e0e12
is an idempotent of C0(q)
such that '(~e) = (1000). Thus
e = ~ee3 + 1
2=
1 + e3 + e0e1 + e0e1e34
is an idempotent of C(q) such that
'(e) =
0BBB@1 0 0 00 0 0 00 0 0 00 0 0 0
1CCCA and '(C(q)e) =
0BBB@IR 0 0 0IR 0 0 0IR 0 0 0IR 0 0 0
1CCCA :Therefore C(q)e is an explicitly given space of spinors for q. As in Example 8, '
induces isomorphisms Spin(q) ~!SL2(IC) and so(q) ~!sl2(IC).
Example 9. Let (V; q) = h�1; 1; 1; 1; 1i over IR. The algebra C(q) is called the
Dirac algebra. By Appendix A,
C(q) 'M4(IC) and C0(q) 'M2(IH):
Let fe0; : : : ; e4g be an orthogonal basis such that q(e0) = �1 and q(ei) = 1; i =
1; : : : ; 4. We decompose q as q1 ? q2 with q1 = h�1; 1i and q2 = h1; 1; 1i, soC(q) ' C(q1) bC(q2). We have, as before, an isomorphism
C(q1) ~!M2(IR); e0 7!
0 1�1 0
!; e1 7!
0 11 1
!
and by Example 6
C(q2) ~!M2(IC); e2 7! �1; e3 7! �2; e4 7! �3;
B. Spinors 129
�i the Pauli matrices. Using Lemma 3 of Chapter 4, we get
: C(q) ~!M4(IC) =M2(M2(IC))
such that
(e0) =
0 1
�1 0
!; (e1) =
0 11 0
!; (e2) =
�1 00 ��1
!;
(e3) =
�2 00 ��2
!and (e4) =
�3 00 ��3
!:
Let a 7! a be the standard involution of M2(IC). The involution
�� :
a bc d
!7! d bc a
!=
0 11 0
! a cb d
! 0 11 0
!
of M2(M2(IC)) is the identity on the images (ei) of ei; i = 0; : : : ; 3. Thus �� is
equal to �0 �1, where �0 is the canonical involution of C(q) (usually we prefer to
work with the standard involution � of the Cli�ord algebra, i.e. the involution which
is �1 on V . But we could not guess how � looks like !).
We now twist by an inner automorphism iu in such a way that the diagram
C(q)iuÆ �! M4(IC) =M2(M2(IC))S S
C0(q)iuÆ �! M2(IH)
is commutative. Let
u =
1 00 �1
!2 GL4(IC);
where �1 = (0110) is the �rst Pauli matrix and let 0 = iu Æ . We get
0(e0) =
0 �1
��1 0
!; 0(e1) =
0 �1�1 0
!; 0(e2) =
�1 00 ��1
!
0(e3) =
�2 00 �2
!and 0(e4) =
�3 00 �3
!:
It follows from Example 6 that 0 maps C0(q) to M2(IH), as claimed. Further the
involution b� = 0�0 0�1 = iu��i�1u is given by b�(x) = u��(u)��(x)(u��(u))�1 or
b� a bc d
!=
0 ��1�1 0
! a cb d
! 0 ��1�1 0
!�1
=
�1d�1 ��1b�1��1c�1 �1a�1
!:
We have �1a�1 = kak�1 for all a 2M2(IC). Thus
b� a bc d
!=
0 �kk 0
! a cb d
! 0 �kk 0
!�1
:
130 B. Spinors
Observe that b� is of symplectic type, as it should be. Since the standard and the
canonical involutions coincide on C0(q), the computation of Spin in Chapter 12
shows that
0(Spin(q)) =�x 2M2(IH)
���� xb�(x) = 1�
=�
a bc d
!2M2(IH)
���� a bc d
! 0 �kk 0
! a cb d
!=
0 �kk 0
!�
so 0 is an isomorphism
Spin(q) ~!SU2(IH; (0k
�k0)):
By formula (12.2) of Chapter 12, we have
Spin(q) ' Spin(�q) ~!SU2(IH; diag(1; �1):
In fact the hermitian matrices (0k�k0) and (�1
001) are congruent since
0 �kk 0
!=
1 1k2�k
2
! 1 00 �1
! 1 �k
2
1 k2
!:
We �nally compute an idempotent of C(q) which generates a simple C(q){module.
We have 0(1+e0e12
) = (1000) 2M2(IC), thus
e = (1 + e0e1
2)(1 + e42
) =1 + e4 + e0e1 + e0e1e4
4
is such that
0(e) =
0BBB@1 0 0 00 0 0 00 0 0 00 0 0 0
1CCCA and '(C(q)e) =
0BBB@IC 0 0 0IC 0 0 0IC 0 0 0IC 0 0 0
1CCCA :
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